Question

josh is driving his 1500kg car at a constant 20.0 m/s. He comes to a top of a hill that is 15.0m tall, if he coasts down the hill:How fast is he ay the bottom? how far up the next hill (300 m high) could his car go if he continues to coast.

Answer 1

At the top of the first hill the initial mechanical energy is:

`\begin{gathered} K_i+U_i=(1)/(2)(1500)(20)^2+(1500)(9.8)(15) \\ =520500 \end{gathered}`

At the bottom the potential energy is zero this means that the final kinetic energy is:

`K_f=520500`

then we have:

`\begin{gathered} (1)/(2)(1500)v^2=520500 \\ v^2=(2\cdot520500)/(1500) \\ v^2=694 \\ v=\sqrt[]{694} \\ v=26.34 \end{gathered}`

Therefore the velocity at the bottom of the first hill is 26.34 m/s

To determine how far up it can go on the next hill we once again use conservation of energy. We know that the initial energy is 520500 J. The final mechanical energy has to be the same but since at the maximum height the velocity is zero then this energy transform in gravitational potential energy, then at the maximum height we have:

`\begin{gathered} mgh=520500 \\ h=(520500)/((9.8)(1500)) \\ h=35.41 \end{gathered}`

Therefore the maximum height the car reaches (if the driver coast it) is 35.41 meters.