Posted by Anonymous on Tuesday, February 16, 2010 at 6:36pm.
Calculate the thermal energy dissipated from brakes in a 1310kg 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.

Physicsenergy  bobpursley, Tuesday, February 16, 2010 at 6:43pm
thermal energy dissipated= change in KE+change in PE
= 1/2 m(vi^2vf^2)+mg520*sin13.5
check my thinking.

Physicsenergy  Damon, Tuesday, February 16, 2010 at 6:47pm
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 (8834)^2 (10^3 m/km/3600s/hr)^2
thermal energy + change in potential energy + change in kinetic energy = 0
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