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Energy & Momentum (Unit Review)

Key Vocabulary Potential Momentum Law of Conservation of Momentum Entropy Law of Machine Elastic Renewable Energy Nonrenewable Energy Mechanical Advantage Mechanical Efficiency

1. Explain what at Joule and a Watt are in terms of their derived units. Explain why this makes sense for each. The unit for joule is (kg * m2)/s2. It is a measure of It is equal to the energy transferred (or work done) to an object when a of one acts on that object in the direction of its through a distance of one meter. A newton is expressed in (kg * m)/s2 or kg (a measure of mass), and m/s2 a unit of which is defined as change in (m/s) per (s). Velocity is a measure of distance (m) per time (s).

The unit for watt is (kg * m2)/s3. It is equal to the work done per time. Work is expressed in joules (a unit described above) per time (s).

2. How is power related to work? Power is the rate of doing work. See derived unit explanation above.

3. List the major types of simple machines and how they work. (HINT: We talked about six) See last page of review packet.

4. Compare and contrast elastic and inelastic with respect to the law of conservation of momentum. A perfectly is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in both types of collisions, but one cannot track the kinetic energy through an inelastic collision since some of its energy is converted to other forms of energy.

5. Define impulse. How is it related to inertia and momentum? Impulse is defined as the product of force and time. Impulse changes the momentum of an object. As a result, a large force applied for a short period of time can produce the same momentum change as a small force applied for a long period of time. Force is a measure of the acceleration of some inertial mass. Likewise change in momentum is a measure of the inertial mass the change is velocity. Therefor change in momentum and impulse are related by their inertial mass and an acceleration (change in velocity).

6. Compare and contrast simple and complex machines. A complex machine is more than one simple machine working together, instead of working alone. Simple machines are (as the name would suggest) simple and include the following: • inclined plane • wheel and axle • lever • pulley • wedge • screw

7. Which of the following statements are TRUE about momentum? a) Momentum is a vector quantity. True b) The standard unit on momentum is the Joule. False c) An object with mass will have momentum. False d) An object which is moving at a constant has momentum. True e) An object can be traveling eastward and slowing down; its momentum is westward. False f) Momentum is a ; the momentum of an object is never changed. False g) The momentum of an object varies directly with the speed of the object. True h) Two objects of different mass are moving at the same speed; the more massive object will have the greatest momentum. True i) A less massive object can never have more momentum than a more massive object. False j) Two identical objects are moving in opposite directions at the same speed. The forward moving object will have the greatest momentum. False k) An object with a changing speed will have a changing momentum. True

8. Which of the following statements are TRUE about impulse? a) Impulse is a force. False b) Impulse is a vector quantity. True c) An object which is traveling east would experience a westward directed impulse in a collision. False d) Objects involved in collisions encounter impulses. True e) The Newton is the unit for impulse. False f) The kg•m/s is equivalent to the units on impulse. True g) An object which experiences a net impulse will definitely experience a momentum change. True h) In a collision, the net impulse experienced by an object is equal to its momentum change. True i) A force of 100 N acting for 0.1 seconds would provide an equivalent impulse as a force of 5 N acting for 2.0 seconds. True

9. Which of the following statements are TRUE about collisions? a) Two colliding objects will exert equal upon each other even if their mass is significantly different. True b) During a collision, an object always encounters an impulse and a change in momentum. True c) During a collision, the impulse which an object experiences is equal to its velocity change. False d) The velocity change of two respective objects involved in a collision will always be equal. False e) While individual objects may change their velocity during a collision, the overall or total velocity of the colliding objects is conserved. False f) In a collision, the two colliding objects could have different acceleration values. True g) In a collision between two objects of identical mass, the acceleration values could be different. False h) Total momentum is always conserved between any two objects involved in a collision. False i) When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. False j) When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change. False k) A moving object collides with a stationary object; the stationary object has significantly less mass. The stationary object encounters the greater collision force. False l) A moving object collides with a stationary object; the stationary object has significantly less mass. The stationary object encounters the greater momentum change. False

10. Which of the following statements are TRUE about work? Include all that apply. a) Work is a form of energy. True b) A Watt is the standard metric unit of work. False c) Units of work would be equivalent to a Newton times a meter. True d) A kg•m2/s2 would be a unit of work. True e) Work is a time-based quantity; it is dependent upon how fast a force displaces an object. False f) Superman applies a force on a truck to prevent it from moving down a hill. This is an example of work being done. False g) An upward force is applied to a bucket as it is carried 20 m across the yard. This is an example of work being done. False h) A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop of the Shockwave ride. This is an example of work being done. True i) The force of acts upon a softball player as she makes a headfirst dive into third base. This is an example of work being done. True j) An eraser is tied to a string; a person holds the string and applies a tension force as the eraser is moved in a circle at constant speed. This is an example of work being done. False k) A force acts upon an object to push the object along a surface at constant speed. By itself, this force must NOT be doing any work upon the object. True l) A force acts upon an object at a 90-degree angle to the direction that it is moving. This force is doing negative work upon the object. False m) An individual force does NOT do positive work upon an object if the object is moving at constant speed. False n) An object is moving to the right. A force acts leftward upon it. This force is doing negative work. True o) A non- is doing work on an object; it is the only force doing work. Therefore, the object will either gain or lose mechanical energy. True

11. Which of the following statements are true about power? Include all that apply. a) Power is a time-based quantity. True b) Power refers to how fast work is done upon an object. True c) Powerful people or powerful machines are simply people or machines which always do a lot of work. False d) A force is exerted on an object to move it at a constant speed. The power delivered by this force is the magnitude of the force multiplied by the speed of the object. True e) The standard metric unit of power is the Watt. True f) If person A and person B do the same job but person B does it faster, then person A does more work but person B has more power. False g) The Newton•meter is a unit of power. False h) A 60-kg boy runs up a 2.0 meter staircase in 1.5 seconds. His power is approximately 80 Watt. False i) A 300-Newton force is applied to a skier to her up a ski hill at a constant speed of 1.5 m/s. The power delivered by the tow rope is 450 Watts. True

12. Which of the following statements are true about kinetic energy? Include all that apply. a) Kinetic energy is the form of mechanical energy which depends upon the of an object. False b) If an object is at rest, then it does not have any kinetic energy. True c) If an object is on the ground, then it does not have any kinetic energy. False d) The kinetic energy of an object is dependent upon the weight and the speed of an object. False e) Faster moving objects always have a greater kinetic energy. False f) More massive objects always have a greater kinetic energy. False g) Kinetic energy is a scalar quantity. True h) An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic energy would be 80 J. True i) An object has a kinetic energy of 40 J. If its speed were twice as much, then its kinetic energy would be 80 J. False j) Object A has a mass of 1 kg and a speed of 2 m/s. Object B has a mass of 2 kg and a speed of 1 m/s. Objects A and B have the same kinetic energy. False k) An object can never have a negative kinetic energy. True l) A falling object always gains kinetic energy as it falls. False m) A 1-kg object is accelerated from rest to a speed of 2.0 m/s. This object gains 4.0 Joules of kinetic energy. False n) If work is done on an object by a non-conservative force, then the object will either gain or lose kinetic energy. False

13. Which of the following statements are true about ? Include all that apply. a) Moving objects cannot have potential energy. False b) Potential energy is the energy stored in an object due to its position. True c) Both gravitational and elastic potential energy are dependent upon the mass of an object. False d) The gravitational potential energy of an object is dependent upon the mass of the object. True e) If the mass of an elevated object is doubled, then its gravitational potential energy will be doubled as well. True f) Gravitational potential energy is lost as objects free-fall to the ground. True g) The higher that an object is, the more potential energy which it will have. True h) The unit of measurement for potential energy is the Joule. True i) A 1-kg mass at a height of 1 meter has a potential energy of 1 Joule. False j) A 1-kg object falls from a height of 10 m to a height of 6 m. The final potential energy of the object is approximately 40 J. False k) If work is done on an object by a non-conservative force, then the object will either gain or lose potential energy. False

14. Which of the following objects have momentum? Include all that apply. a) An is orbiting the nucleus of an . Yes b) A UPS truck is stopped in front of the school building. No c) A Yugo (a compact car) is moving with a constant speed. Yes d) A small flea walking with constant speed across Fido's back. Yes e) The high school building rests in the middle of town. No 15. Consider a karate expert. During a talent show, she executes a swift blow to a cement block and breaks it with her bare hand. During the collision between her hand and the block, the _D_. a) time of impact on both the block and the expert's hand is the same b) force on both the block and the expert's hand have the same magnitude c) impulse on both the block and the expert's hand have the same magnitude d) all of the above. e) none of the above.

16. In order to catch a ball, a baseball player naturally moves his or her hand backward in the direction of the ball's motion once the ball contacts the hand. This habit causes the force of impact on the players hand to be reduced in size principally because _C_. a) the resulting impact velocity is lessened b) the momentum change is decreased c) the time of impact is increased d) the time of impact is decreased e) none of these

17. Suppose that Dirty Harry fires a bullet from a gun. The speed of the bullet leaving the muzzle will be the same as the speed of the recoiling gun _D_. a) because momentum is conserved b) because velocity is conserved c) because both velocity and momentum are conserved d) only if the mass of the bullet equals the mass of the gun e) none of these

18. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount of ____, but different amounts of ____. Pick the two words which fill in the blanks in their respective order. D a. energy, work b. power, work c. work, energy d. work, power e. power, energy f. force, work g. power, force h. none of these 19. A 50.0 kg crate is lifted to a height of 2.0 meters in the same time as a 25.0 kg crate is lifted to a height of 4 meters. The rate at which energy is used (i.e., power) in raising the 50.0 kg crate is _C_ as the rate at which energy is used to the 25.0 kg crate. a. twice as much b. half as much c. the same 20. Suppose that you're driving down the highway and a moth crashes into the windshield of your car. Which undergoes the greater change is momentum? _C_ The greater force? _C_ The greater impulse? _C_ The greater acceleration? _A_ a) a. the moth b) b. your car c) c. both the same

21. In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward at 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they (attach together) and roll together with a speed of _A_. Ignore resistive forces. a. 0.5 m/s b. 0.33 m/s c. 0.67 m/s d. 1.0 m/s e. none of these

22. The firing of a bullet by a rifle causes the rifle to backwards. The speed of the rifle's recoil is smaller than the bullet's forward speed because the _C_. a. force against the rifle is relatively small b. speed is mainly concentrated in the bullet c. rifle has lots of mass d. momentum of the rifle is unchanged e. none of these 23. An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow and string. When released, the arrow will have a kinetic energy of _50_ Joules.

24. A child lifts a box up from the floor. The child then carries the box with constant speed to the other side of the room and puts the box down. How much work does he do on the box while walking across the floor at constant speed? 0 J

25. A 1000-kg car is moving at 40.0 km/hr when the driver slams on the brakes and skids to a stop (with locked brakes) over a distance of 20.0 meters. How far will the car skid with locked brakes if it is traveling at 120. km/hr? a. 20.0 m b. 60.0 m c. 90.0 m d. 120. m e. 180. m 26. A platform diver weighs 500 N. She steps off a diving board that is elevated to a height of 10 meters above the water. The diver will possess _D_ Joules of kinetic energy when she hits the water.

a. 10 b. 500 c. 510 d. 5000 e. more than 5000 .

27. A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which the diver dove was approximately _10_ meters. (HINT: Assume g=10 m/s2)

28. A 5-N force is applied to a 3-kg ball to change its velocity from +9 m/s to +3 m/s. This impulse causes the momentum change of the ball to be _-18_ kg•m/s. The impulse experienced by the ball is _-18_ N•s. The impulse is encountered by the ball for a time of _3.6_ seconds.

Remember impulse is equal to change in momentum.

29. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of 2M in a time of (1/2)t is _4F (4 times force)_.

F = m*(Δv)/t

30. Consider the head-on collision between a lady bug and the windshield of a high speed bus. Which of the following statements are true? List all that apply. D and E a) The magnitude of the force encountered by the bug is greater than that of the bus. b) The magnitude of the impulse encountered by the bug is greater than that of the bus. c) The magnitude of the momentum change encountered by the bug is greater than that of the bus. d) The magnitude of the velocity change encountered by the bug is greater than that of the bus. e) The magnitude of the acceleration encountered by the bug is greater than that of the bus.

31. Rank these four objects in increasing order of kinetic energy, beginning with the smallest. Then in increasing order of potential energy, beginning with the smallest. Object A Object B Object C Object D m = 5.0 kg m = 10.0 kg m = 1.0 kg m = 5.0 kg v = 4.0 m/s v = 2.0 m/s v = 5.0 m/s v = 2.0 m/s h = 2.0 m h = 3.00 m h = 5.0 m h = 4.0 m

KE: D, C, B, A PE: C, A, D, B Object A: KE = 0.5•(5.0 kg)•(4.0 m/s)2 = 40. J PE = (5.0 kg)•(~10 m/s2)•(2.0 m) = ~100 J Object B: KE = 0.5•(10.0 kg)•(2.0 m/s)2 = 20. J PE = (10.0 kg)•(~10 m/s2)•(3.00 m) = ~300 J Object C: KE = 0.5•(1.0 kg)•(5.0 m/s)2 = ~13 J (12.5 J) PE = (1.0 kg)•(~10 m/s2)•(5.0 m) = ~50 J Object D: KE = 0.5•(5.0 kg)•(2.0 m/s)2 = 10. J PE = (5.0 kg)•(~10 m/s2)•(4.0 m) = ~200 J

32. Consider the following physical situations. Identify whether the indicated force (in boldface type) does positive work, negative work or no work.

Description of Physical Situation +, -, or no Work a) a. A cable is attached to a bucket and the force of tension is used to pull the ______+______bucket out of a well. b) b. Rusty Nales uses a hammer to exert an applied force upon a stubborn ______+______nail to drive it into the wall. c) c. Near the end of the Shockwave ride, a braking system exerts an applied ______-______force upon the coaster car to bring it to a stop. d) d. The force of friction acts upon a baseball player as he slides into third ______-______base. e) e. A busy hangs motionless from a silk thread, supported by the ______0______tension in the thread. f) f. In baseball, the catcher exerts an abrupt applied force upon the ball to ______-______stop it in the catcher's mitt. g) g. In a physics lab, an applied force is exerted parallel to a plane inclined at ______-______30-degrees in order to displace a cart up the incline.

h) h. A bob swings from its highest position to its lowest position ______+______under the influence of the force of . 33. Answer the following questions about the scenario below:

System Momentum Before Collision: ______30 kg * m/s, right______System Momentum After Collision: ______30 kg * m/s, right______

Is momentum conserved? ______Yes______Net External Impulse During Collision: ______0 N______

34. Answer the following questions about the scenario below:

System Momentum Before Collision: _____50 kg * m/s, right______System Momentum After Collision: _____45 kg * m/s, right______Is momentum conserved? _____No______Net External Impulse During Collision: _____5 N * s, left______

35. Answer the following questions about the scenario below:

System Momentum Before Collision: ______8 kg * m/s, right______System Momentum After Collision: ______10 kg * m/s, right______Is momentum conserved? ______No______Net External Impulse During Collision: ______2 N * s, right______

36. Determine the unknown velocity in each example. Assume that the collisions occur in an isolated system a) v = 6 m/s

b) v = 11.3 m/s

37. Determine the total kinetic energy of the system before and after the collision and identify the collision as being either perfectly elastic, partially inelastic/elastic or perfectly inelastic. a) KE Before: 50 J; KE After: 50 J; Perfectly Elastic

b) KE Before: 36 J; KE After: 12 J; Perfectly Inelastic

c) KE Before: 183 J; KE After: 153 J; Partially Elastic, Partially Inelastic

38. A 46-gram tennis ball is launched from a 1.35-kg homemade cannon. If the cannon with a speed of 2.1 m/s, determine the muzzle speed of the tennis ball.

Given: mball = 46 g = 0.046 kg; mcannon = 1.35 kg; vcannon = -2.1 m/s

Find: vball = ??? The ball is in the cannon and both objects are initially at rest. The total system momentum is initially 0. After the explosion, the total system momentum must also be 0. Thus, the cannon's backward momentum must be equal to the ball's forward momentum.

vball = 62 m/s

39. Two children are playing with a large snowball while on ice skates on a frozen pond. The 33-kg child tosses the 5.0-kg snowball, imparting a horizontal speed of 5.0 m/s to it. The 33-kg child is 4.0 meters from a 28-kg child and 8.0 meters from the edge of the pond (located behind him). Assuming negligible friction, how much time elapses between when the 28-kg child gets hit by the snowball and when the 33-kg child reaches the edge of the pond? For the sake of the discussion, we will refer to the 33-kg child as the "thrower" and the 28-kg child as the "catcher."

In this scenario, the thrower tosses a snowball forward towards the catcher. This throwing involves an impulse imparted to the snowball. And due to action-, there is an identical impulse imparted to the thrower which causes the thrower to be set in motion in the opposite direction. The impulse is equal to the momentum change. And since the mass and the velocity change of the snowball are known, the momentum change of the snowball can be determined.

m • (Delta v)snowball = m • (vfinal-snowball - vinitial-snowball)

m • (Delta v)snowball = (5.0 kg) • (+ 5.0 m/s - 0 m/s) = 25.0 kg•m/s This 25-unit momentum change of the snowball is equal to the thrower's momentum change. Thus the thrower is moving backwards towards the edge of the pond with a momentum of -25.0 kg•m/s. Since momentum is related to velocity, the post-impulse velocity can be determined.

pthrower = mthrower • vthrower

vthrower = pthrower / mthrower

vthrower = (-25.0 kg•m/s) / (33 kg)

vthrower = -0.7576 m/s The thrower began 8.0 meters from the edge of the pond. Once the ball has been thrown, the thrower is moving backwards towards the edge of the pond with a speed of 0.7576 m/s. Assuming negligible friction on the icy pond, the speed can be used to determine the time that elapses between when the ball is thrown and when the thrower reaches the pond's edge. v = d/t t = d / v

tfor thrower to reach pond's edge = (8.0 meters) / (0.7576 m/s)

tfor thrower to reach pond's edge = 10.56 seconds Once the ball is thrown, the thrower starts moving backwards towards the pond's edge. Meanwhile, the ball is moving forward towards the catcher. The time for the ball to move from the thrower's original position to the catcher is dependent upon the ball's speed and the original distance between the thrower and the catcher. v = d/t t = d / v

tfor ball to move from thrower to catcher = (4.0 meters) / (5.0 m/s)

tfor ball to move from thrower to catcher = 0.800 seconds The instant the ball is thrown, two occur - the ball moves forward towards the catcher and the thrower moves backwards towards the pond's edge. The ball reaches the catcher in 0.800 seconds. And 9.76 seconds (~9.8 s) later (10.56 s - 0.80 s), the thrower reaches the pond's edge.

40. A 9230-kg truck collides head on with a 1250-kg parked car. The vehicles entangle together and slide a linear distance of 10.6 meters before coming to rest. Assuming a uniform coefficient of friction of 0.820 between the road surface and the vehicles, determine the pre-collision speed of the truck. (Note: This is my fault, we didn’t go over coefficient of friction, use a friction acceleration of -8.036 m/s/s; you will also need to use a equation.)

Given: mTruck = 9320 kg; mCar = 1250 kg; vCar-before = 0 cm/s

Find: vTruck-after = vCar-after= ??

Now a kinematic equation can be used to solve for the velocity of the car and truck immediately after the collision. This is shown below:

2 2 vf = vo + 2•a•d 2 2 2 2 2 (0 m/s) = vo + 2•(-8.036 m/s/s)•(10.6 m) = vo - 170.36 m /s

2 2 2 2 (0 m/s) = vo - 170.36 m /s

2 2 2 vo = 170.36 m /s

2 2 vo = SQRT(170.36 m /s )

vo = 13.052 m/s (= vTruck-after = vCar-after ) Now that the post-collision velocity of the car and truck are known, expressions for the total system momentum can be written for the before- and after-collision situations. Before Collision: ptotal-before = (9320 kg)•(vTruck-before) + (1250 kg)•(0 cm/s)

After Collision: ptotal-after = (9320 kg)•(13.052 m/s) + (1250 kg)•(13.052 m/s) Assuming momentum conservation, these expressions are set equal to each other and then

algebraically manipulated to solve for the unknown (mB).

(9320 kg)•(vTruck-before) + (1250 kg)•(0 cm/s) = (9320 kg)•(13.052 m/s) + (1250 kg)•(13.052 m/s)

(9320 kg)•(vTruck-before) = (9320 kg)•(13.052 m/s) + (1250 kg)•(13.052 m/s)

(9320 kg)•(vTruck-before) = 137963 kg•m/s

(vTruck-before) = (137963 kg•m/s) / (9320 kg)

vTruck-before = ~14.8 m/s

41. Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1- kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.

The difficulty of this problem lies in the fact that the collision occurs between two objects moving at right angles to each other. Thus, vector principles will have to be combined with momentum principles to arrive at a solution to the problem. The same conservation of momentum principle will be used; but when summing the before momentum values of the two objects, the fact that they are at right angles to each means that they will have to be added using the Pythagorean theorem. The collision is perfectly inelastic with the two skaters moving at the same speed after the collision. For communication sake, the 39.6-kg skater will be referred to as skater A and the 52.1-kg skater will be referred to as skater B. A vector diagram will likely assist in the solution of the problem.

The individual momentum of the two skaters is first determined.

pA = mA • vA = (39.6 kg) • (6.21 m/s, South) = 245.916 kg • m/s, South

pB = mB • B = (52.1 kg) • (4.33 m/s, East ) = 225.593 kg • m/s, East Now the Pythagorean theorem can be used to add these two vectors and thus determine the pre- collision system momentum. The diagram at the right shows the vectors being added in head-to-tail fashion. The resultant is drawn from the tail of the first vector to the head of the last vector. The

resultant is the hypotenuse of a right triangle whose sides are the pA and pB vectors.

psystem = pA + pB (where pA and pB are right angle vectors) 2 2 psystem = SQRT(pA + pB )

2 2 psystem = SQRT[(245.916 kg • m/s) + (225.593 kg • m/s) ] psystem = 333.717 kg • m/s The direction of the this total system momentum vector can be determined by using a trigonometric function. As shown in the diagram above, the angle theta is the angle between the system momentum vector and the vertical. This angle can be determined using either the tangent, cosine or sine function. The tangent function is used below. tangent(Theta) = opposite side / adjacent side

tangent(Theta) = pB / pA tangent(Theta) = (225.593 kg • m/s) / (245.916 kg • m/s) tangent(Theta) = 0.91735

Theta = tan-1 (0.91735) Theta = 42.532 degrees Before the collision, the total system momentum is 333.717 kg • m/s in a direction of 42.532 degrees east of south. Since total system momentum is conserved, the after-collision momentum of the system is also 333.717 kg • m/s in a direction of 42.532 degrees east of south. After the collision, the two objects move together as a single unit with the same velocity. The velocity of each object can be found by dividing the total momentum by the total mass.

psystem = msystem • vsystem

vsystem = ( psystem ) / (msystem )

vsystem = (333.717 kg • m/s) / (91.7 kg)

vsystem = 3.64 m/s at 42.5 degrees east of south

42. A 1250-kg small aircraft decelerates along the runway from 36.6 m/s to 6.80 m/s in 5.10 seconds. Determine the average resistive force acting upon the plane. (Assume that its engine/propeller makes no contributes to its forward motion).

Given:

vi = 36.6 m/s, vf = 6.80 m/s, and t = 5.10 s

The acceleration of the object is the velocity change per time: a = Delta v / t = (6.80 m/s - 36.6 m/s) / (5.10 s) = -5.84 m/s/s or 5.84 m/s/s, left. This acceleration can be used to determine the net force:

3 Fnet = m•a = (1250 kg) • (5.84 m/s/s, left) = 7.30 x 10 N, left The friction forces (surface and air) provide this net force and are equal in magnitude to this net force.

43. A tow truck exerts a 18300-N force upon a 1210-kg car to drag it out of a mud puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane of motion of the car is horizontal. Determine the time required to drag the car a distance of 6.90 meters from its rest position.

Upon neglecting air resistance, there are four forces acting upon the object. The up and down forces balance each other. The acceleration is rightward (or in the direction of the applied force) since the rightward applied force is greater than the leftward friction force. The horizontal forces can be summed as vectors in order to determine the net force.

Fnet = ·Fx = 18300 N, right - 17900 N, left = 400 N, right The acceleration of the object can be computed using Newton's second law.

ax = ·Fx / m = (400 N, down) / (1210 kg) = 0.3306 m/s/s, right This acceleration value can be combined with other known kinematic information (vi = 0 m/s, d = 6.90 m) to determine the time required to drag the car a distance of 6.9 m. The following kinematic equation is used; substitution and algebra steps are shown.

2 d = vi • t + 0.5 •a • t 2 d = vi • t + 0.5 •a • t 6.90 m = 0.5 • (0.3306 m/s/s) • t2 6.90 m / (0.5 • 0.3306 m/s/s ) = t2 41.4 = t2 6.46 s = t

44. A 22.6-N horizontal force is applied to a 0.0710-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds. Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.

a = Fnet / m = (22.6 N, right) / (0.0710 kg) = 318 m/s/s, right The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown.

vf = vi + a•t vf = 0 m/s + (318 m/s/s)•(0.0721 s)

vf = 23.0 m/s

45. A train has a mass of 6.32x104 kg and is moving with a speed of 94.3 km/hr. The engineer applies the brakes which results in a net backward force of 2.43x105 Newtons on the train. The brakes are held for 3.40 seconds. How far (in meters) does the train travel during this time?

There are three (perhaps four) forces acting upon this train. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding train. In the free-body diagram, these two forces are 5 represented by the Ffrict arrow. The value of this resistive force is given as 2.43x10 N. This is the net force since there are no other horizontal forces; it is the force which causes the acceleration of the train. The acceleration value can be determined using Newton's second law of motion.

5 4 a = Fnet / m = (2.43x10 N) / (6.32x10 kg) = 3.84 m/s/s, left This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds.

2 d = vi • t + 0.5 •a • t d = (26.2 m/s) • (3.40 s) + 0.5 • (-3.84 m/s/s) • (3.40 s)2 d = 89.1 m - 22.2 m d = 66.8 m

46. A student applies a force to a cart to pull it up an inclined plane at constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart.

There are two methods of solving this problem. The first method involves using the equation W = F*d*cos(Theta) where F=20.8 N, d=0.636 m, and Theta=0 degrees. (The angle theta represents the angle between the force and the vector; since the force is applied parallel to the incline, the angle is zero.) Substituting and solving yields W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J. The second method is to recognize that the work done in pulling the cart along the incline at constant speed changes the potential energy of the cart. The work done equals the potential energy change. Thus, W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J

47. A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. If it takes 384 s to climb the hill, then determine ... . a) kinetic energy change of the hiker. b) the potential energy change of the hiker. c) the work done upon the hiker. d) the power delivered by the hiker.

a. Delta KE = 0 J b. Delta PE = +21900 J c. W = +21900 J d. P = 57.0 Watts a. The speed of the hiker is constant so there is no change in kinetic energy - 0 J. b. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE

(m*g*hf). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)] and the initial potential energy is 0 J. So Delta PE = +21900 J (rounded from 21888 J). c. The work done upon the hiker can be found using the work-energy theorem. The equation

reduces to Wnc = PEf

(PEi = 0 J since the hiker starts on the ground; and KEi = KEf since the speed is constant; these two terms can be dropped from the equation since they are equal). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)]. So W = +21900 J (rounded from 21888 J). d. The power of the hiker can be found by dividing the work by the time. P = W/t = (21888 J)/(384 s) = 57.0 Watts

48. A 65.8-kg skier accelerates down an icy hill from an original height of 521 meters. Use what you know about the relationship between work and energy to determine the speed at the bottom of the hill if... a) ... no energy is lost or gained due to friction, air resistance and other non-conservative forces. b) ... 1.40*105 J of energy are lost due to external forces. a) Use the work energy theorem:

KEi + PEi + Wnc = KEf + PEf

The PEf can be dropped from the equation since the skier finishes on the ground at zero height. The

KEi can also be dropped since the skier starts from rest. The Wnc term is dropped since it is said that no work is done by non-conservative (external) forces. The equation simplifies to

PEi = KEf The expressions for KE (0.5*m*v2) and PE (m*g*h) can be substituted into the equation:

2 m*g*h = 0.5*m*vf

where m=65.8 kg, h=521 m, g=9.8 m/s/s. Substituting and solving for vf yields 101 m/s. b) This equation can be solved in a similar manner, except that now the Wext term is -140000 J. So the equation becomes

2 m*g*h - 140000 J = 0.5*m*vf Now substituting and solving for vf yields 77.2 m/s. 49. A 510-kg roller coaster car starts at a height of 32.0 m. Assuming negligible energy losses to friction and air resistance, determine the PE, KE, and speed of the car at the various locations (A, B, C, D, and E) along the track.

Location Height (m) PE (J) KE (J) velocity (m/s) Start 32.0 160 000 J 0 J 0 A 28.0 140 000 J 20 000 J 8.9 m/s B 11.0 55 000 J 105 000 J 20.5 m/s C 20.0 100 000 J 60 000 J 15.5 m/s D 5.0 25 000 J 135 000 J 23.2 m/s E 15.0 75 000 J 85 000 J 18.4 m/s F 0 0 J 160 000 J 25.3 m/s

The potential energy for every row can be found using the equation PE = m*g*h where m=510 kg and g = 9.8 m/s/s. In the first row, the total mechanical energy (KE + PE) equals 160 000 J (rounded). Since no work is done by non-conservative forces, the total mechanical energy must be 160 000 J in all the other rows. So the KE can be computed by subtracting the PE from 160 000 J. The velocity can be found using the equation:

KE = 0.5*m*v2 where m=510 kg. The algebraic rearrangement of this equation results in v = SQRT(2*KE/m).

50. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of 39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a horizontal distance of 25.1 cm.

This is an example of work being done by a non-conservative force (the applied force of the mitt) upon a baseball in order to change its kinetic energy. So

Wnc = Change in KE 2 The change in kinetic energy can be computed by subtracting the initial value (0.5 • m • vi ) from the final value (0 J) .

2 Change in KE = KEf - KEi = 0 J - 0.5 • (0.163 kg) • (39.8 m/s) = -129 J The force can be determined by setting this value equal to the work and using the expression for work into the equation:

Wnc = -129 J F • d • cos(theta) = -129 J F • (0.251 m) • cos(180 deg) = -129 J F = 514 N

51. A 29.1-kg sledder is traveling along a level with a speed of 8.96 m/s when she approaches a gentle incline which makes an angle of 12.5 degrees with the horizontal. If the coefficient of friction between the sled and the incline is 0.109, then what will be her speed at the bottom of the inclined plane, located 8.21 m above the of the incline?

This problem is similar to the above. During the entire descent down the hill, gravity is doing work on the sledder and friction is doing negative work on the sledder. Friction is a non-conservative force and will alter the total mechanical energy of the sledder. The equation to be used is

KEi + PEi + Wnc = KEf + PEf If we designate the level area at the bottom of the slope as the zero level of potential energy, then

PEf is 0 J. So the above equation becomes KEi + PEi + Wnc = KEf 2 2 0.5•m•vi + m•g•hi + F•d•cos(theta) = 0.5•m•vf

The Wnc term results from friction acting upon the sledder. The force of friction

is given by the equation Ffrict = mu • Fnorm. For inclined planes the normal force is equal to the perpendicular component of the weight vector (see diagram at the right). This component is equal to the expression m•g•cos(theta) where theta is the incline angle (12.5 degrees). So the force of friction is

Ffrict = mu•m•g•cos(theta)

Ffrict = (0.109)•(29.1 kg)•(9.8 m/s/s)•cos(12.5 deg) = 30.3 N The work done by the friction force occurs over the entire length of the inclined plane. This distance

is related to the initial height and the incline angle by the equation sin(theta) = hi/d (see diagram below). Using 12.5 degrees for theta and 8.21 m for the initial height, the distance along the incline (d) can be determined to be 37.9 m. The diagram below depicts the physical situation.

Now substitutions can be made into the earlier stated equation and algebra manipulations can be performed to determine the final speed. The work is shown below.

2 2 0.5•m•vi + m•g•hi + F•d•cos(theta) = 0.5•m•vf 0.5•(29.1 kg)•(8.96 m/s)2 + (29.1 kg)•(9.8 m/s/s)•(8.21 m) + (30.3 N)(37.9 m)•cos(180 deg)= 2 0.5•(29.1 kg)•vf

2 1168 J + 2341 J - 1148 J = (14.6 kg)•vf

2 2 2 162.3 m /s = vf

2 2 vf = SQRT(162.3 m /s ) = 12.7 m/s

52. Suzie Lovtaski has a mass of 49.7 kg. She is on her skateboard at rest on top of a hill with a height of 92.6 m and an incline angle of 19.2 degrees. She coasts down the hill to the bottom and eventually comes to a stop. The coefficient of friction is 0.0873 along the hill and 0.527 along the horizontal surface at the bottom. What total distance will Suzie coast (include both incline and level surface)?

There is a non-conservative force - friction - doing work upon the skateboarder. This force will alter the total mechanical energy of the skier. The equation to be used is

KEi + PEi + Wnc = KEf + PEf If we designate the level area at the bottom of the slope as the zero level of potential energy, then

PEf is 0 J. Since Suzie eventually stops (due to the effect of friction along the level area), the KEf is

0 J. And since Suzie starts from rest, KEi = 0 J. So the above equation becomes

PEi + Wnc = 0

The Wnc term has two parts; there is friction doing along the inclined plane and friction doing work along the level surface. Since these two sections of the motion have different normal forces and friction coefficients (and therefore friction forces), they will have to be treated separately. The graphic to the right depicts the free-body diagrams and the means by which the friction force can be determined.

By substituting values of mu and mass and g and theta into the above equations, one finds that the friction values are

On Incline On Level Surface

Ffrict = 40.2 N Ffrict = 257 N These forces act upon the skier over different distances. In the case of the inclined plane, the distance (d) can be computed from the given incline angle and the initial height. The relationship is depicted in the diagram below. The sine function is used to relate the angle to the initial height and the distance along the incline. In the case of the level surface, the distance is the unknown quantity (x) which this problem calls for. The distance d along the incline is

d = hi / sin(theta) = 92.6 m / sin(19.2 deg) = 282 m

Now substitutions can be made into the work-energy equation and algebraic manipulation can be performed to solve for x:

PEi + Wnc = 0

PEi + Wincline + Wlevel = 0 (49.7 kg)•(9.8 m/s)•(92.6 m) + (40.2 N)•(282 m)•cos(180 deg) + (257 N)•(x)•cos(180 deg) = 0 J 45102 J - 11307 J - 257 x = 0 J 33795 J = 257 x 132 m = x