question 4 or 5 1. A frictionless mass is at the bottom of a hill, as shown belo

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question 4 or 5

1. A frictionless mass is at the bottom of a hill, as shown below. The mass can be pushed against a spring so that when the mass is released, it goes a height h up the hill. What spring compression r is necessary for the mass to reach a given height h? Im 2. A child of mass m-20 kg goes down a slide at a playground. She starts at a height h- 2.2 m with zero velocity. Assuming zero friction, how fast will she be going at the bottom? 3. When you do the experiment in the previous problem, you measure the child's final speed to be 1.8 m/s, which is lower than the answer you might expect. Explain where the "missing" energy went, describe the other force that did work on the child, and calculate how much work that force did 4. One problem Tarzan had to deal with while swinging on vines is that vines grow on trees. He started swinging one one such vine (L14 ) with an initial angle of 020 measured from vertical. When he reached an angle 0f+4°, he met a tree face-first. How fast was he going? Assume that Tarzan's mass was 80 kg. 5. One area of human endeavor that would seem to require a thorough understanding of conserva- tion of energy is bungie jumping. You have a bungie cord with a spring constant k 700 N/m and an unstretched length Lo 40 . Your mass is (for this problem) m 70 kg. How high a bridge do you need! Hint: Use conservation of mechanical energy, obviously, but remember that although you gain kinetic energy for much of the way down, and you lose gravitational potential energy all the way down, you only gain spring potential energy after you have fallen a distance Lo. 6. A 60-kg backpacker with a 15-kg pack is hiking out of the Grand Canyon along the Bright Angel Trail. The total elevation change is 1340 m. (a) What will be her change in potential energy during this hike? (b) If she makes it out in 3 hours, what power will that require? 7. Bubba punts a Zero-DragTN football off the top of the physical science building at an angle above the horizontal, with speed u-27 m/s. How fast will this football be moving when it hits the ground 10 m below?

Explanation / Answer

4. Applying energy conservation,

PEi + KEi = PEf + KEf

- m g L cos20 + 0 = mv^2 /2 - m g L cos4

v = sqrt(2 x 9.8 x 14 ( cos4 - cos20))

v = 3.98 m/s ......Ans

5. Applying Work - energy theorem,

Work done by gravity + work done by spring = change in KE

70 x 9.8 x (40 + h ) - 700 h^2 /2 = 0 - 0

350 h^2 - 686 h - 27440 = 0

h = 9.9 m

hence H = 40 + h = 49.9 m

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