Calculate the thermal energy dissipated from brakes in a 1310-kg car that descends a 13.5° hill. The car begins braking when its speed is 88 km/h and slows down to a speed of 34 km/h in a distance of 0.52 km measured along the road.

Answer 1

thermal energy dissipated= change in KE+change in PE

= 1/2 m(vi^2-vf^2)+mg520*sin13.5

check my thinking.

Answer 2

work done = force * distance

F = m a

Change in potential energy = - m g (.52*10^3) sin 13.5

Change in kinetic energy = - (1/2) m (88-34)^2 (10^3 m/km/3600s/hr)^2

thermal energy + change in potential energy + change in kinetic energy = 0