A cart starts at rest and moves freely down a ramp without friction or air resistance and descends 8 meters vertically, what is its speed at the bottom?

We apply energy conservation. At the start the cart has only gravitational potential energy given by mgh where m is its mass, g is the gravitational field, h = 8m is the height. At the end the cart has only kinetic energy mv^2/2 where v is its speed. By conservation of energy mgh = mv^2/2, so v^2=2gh= 29.88 m^2 s^-2=156.8 m^2 s^-2 so taking the square root, v = 12.5 m s^-1.

JT
Answered by Joshua T. Physics tutor

1511 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

What is the minimum initial velocity necessary for an object to leave Earth?


Please see below.


A 100g mass is on a circular turntable spinning at 78 revolutions per minute. The maximum frictional force between the mass and turntable is 0.50N. Find the maximum distance from the center of the turntable at which the mass would stay on the turntable.


A pendulum of mass m is released from height h with a speed v at the bottom of its swing. a) What is the gravitational potential energy at height h and the kinetic energy at the bottom of its swing? b) Use conservation of energy to define the speed v.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning