Question

A 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reaches the bottom of the hill? (Assume g = 9.81 m/s2.)

Answer 1

Considering conservation of mechanical energy we have:
K+U (start)= K+U (end)
Where:
K=kinetic energy=$\frac{1}{2}m v^{2}$
U=potential energy=$mgh$
So:
0.05*0.95*9.81+0 (start) = 0+K (end)
K (end)=0.5 Joules