Question

A 10.0 kg box is slide up an 25-deg incline for a total distance of 2.5 m by applying 3000-N of force directed up the incline. The coefficient of friction between the box and the plane is 0.16. (a) What is the acceleration up the plane? (b) How much work was done by friction? (b) How much work was done by the applied force?

Answer 1

Here,

m = 10 Kg

theta = 25 degree

L = 2.5 m

F = 3000 N

a) acceleration up the incline = (F - m * g * sin(theta) - u *m g * cos(theta))/m

acceleration up the incline = (3000 - 10 * 9.8 * sin(25 degree) - 0.16 * 10 * 9.8 * cos(25 degree))/10

acceleration up the incline = 294.4 m/s^2

b)

work done by friction= u *m g * cos(theta) * L

work done by friction= 0.16 * 10 * 9.8 * cos(25 degree) * 2.5

work done by friction= -35.5 J

c) work done by applied force = 3000 * 2.50

work done by applied force = 7500 J