Question

A 1250 kg car drives up a hill that is 15.2 m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car's engine. The work done by friction is −2.91×105 J ; the work done by the engine is 6.44×105 J .

Part A Find the change in the car's kinetic energy from the bottom of the hill to the top of the hill.

Answer 1

Using Work-energy theorem:

W = dE

Work-done by non-conservative force = Wf + We

Wf = Work-done by friction = -2.91*10^5 J

We = Work-done by Engine = 6.44*10^5 J

dE = dKE + dPE

dPE = Change in potential energy = PEf - PEi = m*g*(hf - hi)

dKE = Change in kinetic energy = KEf - KEi = change in KE from the bottom to top of the hill

So, dKE = W - dPE

dKE = Wf + We - m*g*(hf - hi)

hf = 15.2 m, hi = 0 m

m = mass of car = 1250 kg

So,

dKE = -2.91*10^5 + 6.44*10^5 - 1250*9.81*(15.2 - 0)

dKE = 166610 = 1.67*10^5 J

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