Work and Energy

Question

Suppose a car approaches a hill and has an initial speed of 120 km/h at the bottom of the hill. The driver takes her foot off of the gas pedal and allows the car to coast up the hill.

a) if the car has the initial speed stated at a height of h=0, how high in meters can the car coast up a hill if work done by friction is negligible?

b) if in actuality a 710-kg car with an initial speed of 120 km/hr is observed to coast up a hill and stops at a height 21.5m above its starting point, how much thermal energy was generated by friction in J?

c) what is the magnitude of the average force of friction in newtons if the hill has a slope 2.5 degrees above the horizontal?

Answer 1

Part a)

If there is no friction then mechanical energy is conserved. This means that

E=KE+PE is the same at all times. Importantly, we can set energies at different times equal to each other.

So, the energy at the bottom of the hill must be the same as the energy at the end, when it has climbed to its maximum and stopped before rolling back down.

At the beginning, the energy is all kinetic, E=KE=1/2 mv². At the end, the car has reached its maximum height and has lost all its speed. It's dumped all its kinetic energy into potential energy to climb the hill, so to speak. So, E=PE=mgh_max.

Then, we have 1/2 mv²=mgh_max. Solving for h_max, we get h_max=v²/2g. Plugging in numbers after converting the speed they gave us to SI units of m/s, we get about h_max=56.6m.

Part b)

If the car didn't get to 56.6m, we know the remainder of its energy must have gone to friction.

So we started with E_i=1/2 mv²=3.94*10⁵J using the numbers provided, and ended with E_f=mgh=1.48*10⁵J using the achieved height h=21.5m. The difference must have been lost to friction, so the answer is E_i-E_f=2.45*10⁵J , or about 245kJ.

Part c)

The work done by friction will be the same as the energy generated by friction by the work-energy theorem. This work is defined by W=F*d where F is the force doing work and d is the distance over which the work is done. If the curve has a slope of 2.5deg, and I know the height reached is 21.5m, then I can calculate the hypotenuse of this triangle, which is the distance over which friction was being applied. Using some trigonometry, the hypotenuse is d=21.5/(sin(2.5deg))= 493m.

So, the force needed to generate the 245kJ of work over a distance of 492.9m will be

F=W/d=245kJ/0.493km = 496N