Energy Conservation: A 300 – kg cart’s brakes has a velocity of 8.00 m/s on a hill that is 50.0 m high. The cart

rolls down the first hill and up a second hill. It the height of the second hill is 30.0 m,

what is the cart’s velocity on the top of the second hill?

The cart's velocity on the top of the second hill is 20m/s

Explanation:

Mass of the cart, m1 = 300kg

Velocity of the cart, v = 8m/s

Height of the hill, h1 = 50m

Height of the second hill, h2 = 30m

Cart's velocity on the top of the second hill, vf = ?

We have to apply conservation of energy.

Initial Kinetic energy + Initial Potential energy = final kinetic energy + final

                                                                                                  potential energy

Initial kinetic energy is 0 because the cart is at rest.

It will have initial potential energy of mgh acting on it as it is on height.

The final potential energy will be 0 and the final kinetic energy will be 1/2mv²

Thus,

Therefore, the cart's velocity on the top of the second hill is 20m/s

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2