A cart with a mass of 100.0 kg is filled with 1800.0 kg of coal. A force of 14.0

Question

A cart with a mass of 100.0 kg is filled with 1800.0 kg of coal. A force of 14.0 kN is being used to pull the cart of coal up a short, 10.5 m, 23.0 degree incline. The cart experiences friction as it is being pulled. The coefficient of kinetic friction for the cart is 0.300.

a. What is the work done by the force pulling the cart up the incline?

b. What is the work done by gravity?

c. What is the work done by friction?

d. What is the work done by the normal force?

e. If the cart begins at rest, what is the magnitude of the cart's velocity at the top of the hill?

Explanation / Answer

Work done by the force pulling the cart = F * d = 14 * 10^3 * 10.5 = 147000 J

Work done by gravity = - 1900 * 9.8 *10.5*sin(23) = -76391.84 J

Work done by friction = - 0.3 * 1900 *9.8 * 10.5 = -58653 J

Work done by normal force = Zero (as it is perpendicular to the direection of motion)

Total work done = change in kinetic energy

147000 -76391.84-58653 = 0.5 * 1900 * v^2

v = 3.54 m/s

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