A 100 kg box is sliding down a ramp with an acceleration of 3 m/s2.

Question

A 100 kg box is sliding down a ramp with an acceleration of 3 m/s². If the ramp makes an angle of 45 degrees with the ground, find the normal force, the friction force, and the coefficient of friction between the box and the ramp.

Please include:

1. An FBD
2. Complete solutions showing all of your work
3. A written explanation of 1-2 sentences explaining what your numbers mean

Answer 1

friction force = umgcos(theta)

Down incline gravity force = mgsin(theta)

net downward force = mgsin(theta)-umgcos(theta) = ma

u = coefficient of kinetic friction

a= gsin(theta)-ugcos(theta)

Theta = 45

g=10 m/s^2

3=10/sqrt(2)-(u*10)/sqrt(2)

3sqrt(2)=10-10u

10u=10-3sqrt(2)

u= [10-3sqrt(2)]/10

u=0.5757