Cindy is changing the tire on her car on a steep hill, 27.6 m high. She trips and drops the 8.7 kg spare tire, which rolls down the hill with an initial speed of 2.37 m/s. What is the speed of the tire at the top of the next hill, which is 7.9 m high?

Question

Gerardo Mcbride

Physics

13 January, 20:18

Cindy is changing the tire on her car on a steep hill, 27.6 m high. She trips and drops the 8.7 kg spare tire, which rolls down the hill with an initial speed of 2.37 m/s. What is the speed of the tire at the top of the next hill, which is 7.9 m high?

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Mike Jimenez · Answer 1

Initially, the tire has kinetic energy and potential energy. If we assume 0 potential energy at 7.9 m, then the initial energy is

E = mgh + ½mv² = 8.7kg * 9.8m/s² * (27.6 - 7.9) m + ½ * 8.7kg * (2.37m/s) ² = 1,690 J

With on no account of energy loss, this will be the energy at the topmost of the 7.9m hill, which we distinct as having 0 potential energy. Then:

E = 1690 J = ½mv² = ½ * 8.7kg * v² v = 41 m/s