Question

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?​

Answer 1

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

Step-by-step 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,

`0 + mg (h1 - h2) = (1)/(2) m (v)^2 + 0\\\\mg (50 - 30) = (1)/(2) m (v)^2\\\\20g = (1)/(2) (v)^2\\\\20 X 10 X 2 = (v)^2\\\\400 = (v)^2\\\\v = 20m/s`

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