Problem Set #3a & #3b solutions

1a. You must calculate the horizontal component of the force vector because the box’s displacement happens in the horizontal direction. 1b. You must know how far the cart is moving WHILE you are applying the force to it. 1c. You must realize that you are applying a force in one direction and the displacement is happening in the opposite direction, meaning you will have a negative work value.

2. As described above…if you apply force in 1 direction (+Y) and the displacement is in the opposite direction (-Y), you will have a negative work value.

3. Mechanical energy---throwing a ball (units are Joules) : Thermal Energy---energy due to heat (units are Joules) : Chemical energy---energy stored in chemicals (compounds) and energy released or absorbed in chemical reactions (units are Joules).

4. There is 2 (maybe 3) types of Mechanical Energy, 1 type of KE and 2 types of PE.

5. KE = ½ * m * v2 --- 3000 N/9.8 kg*m/s2 = 306.1 kg --- KE(vehicle) = .5 * 306.1 kg * (100 m / 35 s)2 = 1249.5 J --- KE(bullet) = .5 * .01 kg * (1200m/s)2 = 7200 J --- The bullet has more KE.

6. PE = m*g*h --- 4 stories * 10 m = 40 meters --- PE = w*h = 29 N * 40 m = 1160 J

7. You can use projectile motion equations to solve for velocity of the flowerpot just before impact in a 40 m drop. Once you have this velocity you can use the KE equation to solve for KE. You should also know that during a fall like this where gravity is the only force acting that Mechanical Energy is conserved and doesn’t change. Knowing this and that all the potential energy in number six will be transferred to KE at the bottom of the fall tells you that KE = 1160 J. Since work is equal to the change in energy and the energy goes from 1160 J to 0 J…the work required to stop the pot is 1160 J.

8. Work = Force * Displacement & Work = (KE2 + PE2) – (KE1 + PE1) = E

9. Just make up numbers here…For a height of 100 meters and a mass of 100 kg…PE is 100 kg * 9.8 kg*m/s2 * 100 meters = 98000 J. If this object is dropped, you can calculate final velocity and then calculate KE --- Vf2 = Vi2 + 2ad --- Vf = 44.27 m/s --- so KE at ground (just before ground) = ½ * 100 kg * (44.27 m/s)2 = 97991 J (approximately 98000). ME at the top is KE + PE and KE is equal to 0, so ME = 98000 J. At the bottom PE is 0 and KE is 98000 J so total ME is 98000 J. ME is conserved in the absence of external forces excluding gravity.

10. English units (Hp) and SI units (Watts) compare as follows: 1 hp = 745.7 W

11. THIS IS A BAD QUESTION: IT SHOULD SAY TO CALCULATE THE POWER REQUIRED TO LIFT A 500 KG VEHICLE 1050 METERS IN 1 SECOND. Then P = W/t --- F*d/t ---500 kg * 9.8 m/s2 * 1050 m / 1 s --- Power = 5.145 X 106 W 12. SAME HERE AS ABOVE, SHOULD SAY --- CALCULATE THE POWER NECESSARY TO LIFT A 300 KG VEHICLE AT 400 M/S. P = F * V --- 300 kg * 9.8 m/s2 * 400 m/s = 1.176 X 106 W.

13. Impacts a.) happen very quickly b.) involve contact forces of high magnitude c.) result in rapid changes in momentum of one or both colliding objects.

14. A perfectly elastic impact has a coefficient of restitution of 1 and no energy is lost, whereas a perfectly plastic impact has a coefficient of restitution of 0 and all energy is lost, in reality most impacts are somewhere in between.

15. Perfectly elastic = 1, perfectly plastic = 0, most impacts 0 < e < 1.

16. e = -(V1 – V2) / (U1 – U2) --- V1 = 20 and V2 = 20 giving the top of the equation 20 – 20 --- e = 0.

17. 1st use projectile motion to calculate the height the golf ball travels. --- Vf2 = Vi2 + 2ad --- d = 250 meters --- e = sqrt (hb/hd) --- so bounce height is equal to 2.5 meters --- PE is this case (using golf ball mass of 0.046 kg) .046kg * 9.8 m/s2 * 2.5 m = 1.127 J at the top of the first bounce --- half way down from this bounce the total ME will be half KE and half PE --- if the total ME is 1.127 J (at the top 1.127 J PE and 0 J KE, so total ME = 1.127 J) --- the KE half way down will be 1.127 J / 2 = 0.5635 J

Problem Set 3b

18. Forces cause changes in LINEAR motion whereas torques cause changes in ANGULAR motion.

19. T = r X F --- r is the moment arm (perpendicular distance between force and axis of rotation), F is the force, and T is the torque (Turning effect produced by a force)

20. This problem must be a statics problem. M = 0 --- pick a positive direction, probably counter clockwise, then sum torques and set this equal to 0, 50 lbs * 10 inches – F * 1 inches = 0 --- F = 500 lbs.

21. Same technique as above. 13 feet 5 inches = 13.42 ft --- 200 N * 13.42 ft – 140 N * r = 0 --- r = 19.17 ft.

22. 9 in = .2286 m --- T = r X F = 95 N * .2286 m = 21.72 N*m --- when the wrench is at 45 degrees, the perpendicular force is 95 N cos 45 = 67.2 N ---so T = .2286 m * 67.2 N = 15.36 N*m --- when at 30 degrees, F = 95 N cos 30 = 82.3 N --- T = .2286 m * 82.3 N = 18.8 N*m. 23. In events like the high jump or the pole vault this happens when going over the bar is done correctly. If you bend at the wait a lot and get much mass away from your core, you CM can indeed be located outside of your body.

24. INERTIA is the property of an object that resists changes in linear motion, whereas MOMENT OF INERTIA is the property of an object that resists changes in angular motion.

24.5. SHOULD HAVE SAID ROTATING AT 25 RAD/S. H = 300 kg * m2/s (angular momentum) --- H = I *  --- 300 kg * m2/s / 25 rad/s = I = 12 kg * m2 --- if she doubles her radius of gyration, her new moment of inertia will be 4X as much since r is squared. This would make the new moment of inertia I = 48 kg * m2.

25. BAD QUESTION

26. radius of gyration is the distance from the axis of rotation you would have to place all the mass of an object to have it act as it did in angular motion before you centered all the mass. No you would have to smash all your leg into a little ball.

27. Because radius is squared in the equation. In the horizontal range of a projectile velocity is squared, and in KE velocity is squared.

28. The runner reduces the r term in the moment of inertia calculation, which reduces the moment of inertia to allow him/her to angularly accelerate the leg more easily to bring it back forward to get into position to produce force again.

29. L represents linear momentum and H represents angular momentum --- L = m * v --- H = I * 

30. No forces other than gravity act on the diver during the dive. This means that the angular momentum will not change. When the diver tucks, he/she reduces moment of inertia, but since angular momentum cannot change, angular velocity must increase to balance the decrease in moment of inertia. The opposite is true as they untuck towards the end of the dive.

31. T = I * 

32. Impulse is equal to change in momentum, angular impulse is equal to change in angular momentum.

33. In the tightrope, the performer carries a long stick or pole with a very large moment of inertia. This stick does not want to be rotated. If the performer loses his/her balance, they can apply a torque in one direction to the pole and the pole will apply an equal and opposite torque to the performer, allowing them to balance one again. On the balance beam if you lose your balance one way you rotate your limbs hoping the beam will give you an equal and opposite torque and bring you back up on the beam.