Work and Kinetic Energy

Level : Physics I Teacher : Kim

Objective Establish the relationship between work and energy Practice using Work-Kinetic Energy theorem compared to using ΣF=ma Understand how work&energy leads to the concept of simple machinery we use everyday

Observe video clip on ‘Work’

Introduction of ‘Work’ done by Applied Force - Work(W) is defined as ‘force applied in the direction of motion × displacement’

W=Fcosφ × d [J]

Fapp

d= distance traveled

- Both force and distance is directly related to work. - Unit of work is expressed in Joules(J) - φ is the angle the between the direction of force and the direction of motion - If you apply 1N of force over a distance of 1meter, then you are doing 1Joule of work.

Q1) A boy exerts a horizontal force of 10N on a box at a distance of 25m on a flat surface. How much work did he do on the box? a) 150J b) 200J c) 250J d) 300J

Q2) A girl exerts a force of 35N at an angle of 35° above the horizontal along the handle of her wagon as she pulls it 25m across a level sidewalk. How much work does she do on her wagon? a) 698J b) 717J c) 756J d) 782J

Lifting a barbell is doing work. Two things happen when lifting a barbell i) a force is exerted on the barbell, and ii) the barbell is moved by that force Here, work is done on the barbell, since there is force being applied on the barbell and the barbell moves a certain distance

Q3) A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J

Q4) If the man is just holding the barbell above his head, how much work is done on the barbell? a) He is doing no work on the barbell. Since W=F×d and distance covered is zero, so W=0. b) Of course he is doing work, since holding the barbell above his head causes strain on his body

Work done by other Forces - Any force that may exist on an object has the ‘potential’ to do work. In the figure below, three forces are acting on the block; FN, Fg and Fapp. But only Fapp contributes to doing work on the block. (surface is frictionless)

FN Fapp

d= distance traveled Fg=mg

=> Work done by normal force FN is zero; W=FN cos90º × d =0

=> Work done by gravitational force Fg(=weight) is zero; W=Fg cos90º × d = 0

*~ The only time a force does work on an object is when the force ‘plays a role’ in moving the object Work done by Gravity i) Vertical Motion Q5) When an object is free-falling in a vacuum, what force is doing work on that object? Fg ii) Incline Plane Q6) When an object is sliding down on the FN frictionless ramp, what force is doing work on d that object? Fg sinθ

Fg θ

*~Work-Kinetic Energy Theorem

If an object is in motion, then it must have kinetic energy(K)

K= mv2 1

2 If you apply a force over a certain distance, there will be a change in kinetic energy

vi distance ‘d’ vf

Fapp Fapp

Ki + Fcos × d = Kf

𝝋𝝋

2 2 mvi +Fcosφ × d = mvf 1 1 2 2

Ki + W = Kf : Work-Kinetic Energy Theorem

𝜮𝜮

- Unit of energy or work is Joules[J]

*~ Are we interested in the ‘acceleration’ or the ‘speed’??~*

2 W=Fcosφ × d K= mv Ki + W = Kf 1 2 𝛴𝛴 A. On a Flat Surface Q7) If a 5kg object is sliding on a frictionless surface with a constant speed of 10m/s, i) What is the kinetic energy of the object?

ii) If you apply a horizontal force of 5N over a distance of 10m, what is the new kinetic energy? Use ΣF=ma and Work-Kinetic Energy theorem and compare the process and result

Newton’s 2nd Law : ΣF=ma Work-Kinetic Energy theorem 2 2 *~ You need to also use vf -vi =2ad

Q8) A 4kg object is sliding on a frictionless surface with a constant speed of 8m/s, and you apply a horizontal force of 8N over a distance of 12m. i) How much work was done by the applied force? a) 96J b) 82J c) 74J d) 68J

ii) What is the new kinetic energy? a) 162J b) 178J c) 195J d) 224J

iii) What is the new speed? a) 10.6m/s b) 8.4m/s c) 7.1m/s d) 5.9m/s

2 W=Fcosφ × d K= mv Ki + W = Kf 1 2 𝛴𝛴 Q9) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you

apply a horizontal force of Fp=6N, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s

Q10) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you

apply a force of Fp=6N at an angle of 27°above the surface, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s

Fp 27°

Q11) An object of mass 2kg is sliding on a frictionless surface with a constant speed of 8m/s.

How much force Fp is needed to stop the object over a distance of 10m? (*~remember that the force is applied in the opposite direction of the motion for the object to stop~*) a) 3.9N b) 6.4N c) 8.3N d) 11.8N

FN direction of motion Fp

2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 𝛴𝛴 Q12) A 2.5kg box is resting on the flat rough surface. The coefficient of kinetic friction between the box and the floor is µk =0.52.The box is then pulled by a force of Fp=18N.

FN direction of motion fk Fp

Fg i) How much work was done on the box by the pulling force Fp through a distance of 5.2m?

That is, find Wp. a) 62.4J b) -62.4J c) 93.6J d) -93.6J

ii) How much work was done on the box by the frictional force fk through a distance of 5.2m?

That is, find Wf. a) 66.25J b) -66.25J c) 25.48J d) -25.48J

iii) What is the net work W done on the box? That is, Wp + Wf. a) 27.4J b) 33.5J c) 49.7J d) 75.14J 𝛴𝛴

iv) What is the speed of the box when it covered 5.2m? a) 39.8m/s b) 27.2m/s c) 15.6m/s d) 4.7m/s

Q13) A 4kg box is sliding on a rough surface with an initial speed of 8m/s. The coefficient of

kinetic friction between the box and the floor is µk =0.4. What is the speed of the box after it moved a distanced of 4m? (*~frictional force f=µFN, where FN=mg in this case ~*) a) 19.6m/s b) 14.2m/s c) 10.5m/s d) 5.7m/s

FN direction of motion fk

2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 𝛴𝛴 B. Vertical Motion Use Work-Kinetic Energy to solve Q3) again! A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J

C. Incline Plane Q14) A 8kg box is released from rest on a 30° inclined frictionless plane and accelerates down the incline. The distance from the box to the end of the ramp is 4m

i) Find the work done by gravity through a distance of 4m down the ramp. a) 271.6J b) 156.8J c) 101.2J d) 59.7J

ii) What is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s

Q15) A 13.2kg box is released from rest on an 11° inclined frictionless plane and accelerates down the incline. If the distance from the box to the end of the ramp is 5.1m, what is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s

2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 𝛴𝛴 Q16) A 6.3kg box is being pulled with a force of 56N from the bottom of the ramp starting from rest. The surface of the ramp is frictionless and the angle of the ramp is 40°.

i) How much work was done on the box by the pulling force Fp through a distance of 5.2m?

That is, find Wp. a) 291.2J b) 233.9J c) 187.2J d) 105.6J

ii) How much work was done on the box by gravitational force Fg through a distance of 5.2m?

That is, find Wg. a) 245.9 b) -245.9J c) 206.4J d) -206.4J

iii) Find the speed of the box when it reaches a distance of 5.2m from the bottom. a) 26.9m/s b) 13.4m/s c) 5.19m/s d) 1.21m/s

Q17) A 5.6kg box is being pulled with a force of 40N from the bottom of the ramp starting from rest. The surface of the ramp is frictionless and the angle of the ramp is 35°. Find the speed of the box when it reaches a distance of 6.4m from the bottom.

a) 2.56m/s b) 4.41m/s c) 6.78m/s d) 8.67m/s

2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 𝛴𝛴 Q18) A 8kg box is released from rest on a 30° inclined rough plane and accelerates down the incline. The coefficient of friction µk =0.54.

i) If the distance from the block to the end of the ramp is 4m, what is the speed of the block when it reaches the bottom of the ramp? a) 3.3m/s b) 2.4m/s c) 1.6m/s c) 0.4m/s

ii) The block will eventually slide off the ramp and continue on the surface. If the coefficient of kinetic friction on the flat surface was µk =0.12, how far will the block travel until it stops? a) 2.21m b) 1.63m c) 1.07m d) 0.64m

2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 𝛴𝛴 Q19) A 3.6kg box is being pulled with a force of 42N from the bottom of the ramp starting from

rest. The surface is rough, so the coefficient of friction μk=0.13, and the angle of the ramp is 26°. Fp

i) Find the work done by the pulling force, Wp, through a distance of 4.8m. a) 201.6J b) 148.4J c) 96.4J d) 62.2J

ii) Find the work done by the gravitational force Wg, through a distance of 4.8m a) 37.1J b) -37.1J c) 74.2J d) -74.2J

iii) Find the work done by the frictional force, Wf, through a distance of 4.8m a) 19.8J b) -19.8J c) 39.6J d) -39.6J

iii) What is the net work W done on the box? That is, Wp + Wg + Wf. a) 46.5J b) 107.6J c) 168.4J d) 197.5J 𝛴𝛴

iv) Find the speed of the box when it reaches a distance of 4.8m from the bottom. a) 1.23m/s b) 3.18m/s c) 5.54m/s d)7.73m/s