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,
[tex]0 + mg (h1 - h2) = \frac{1}{2} m (v)^2 + 0\\\\mg (50 - 30) = \frac{1}{2} m (v)^2\\\\20g = \frac{1}{2} (v)^2\\\\20 X 10 X 2 = (v)^2\\\\400 = (v)^2\\\\v = 20m/s[/tex]
Therefore, the cart's velocity on the top of the second hill is 20m/s
