friction on an incline plane!?

Question

A 6.0 kg box slides up a 8.0 m long friction incline at a constant speed at an angle of 30 degree by force 50 N parallel to the incline. The coefficient of kinetic friction is 0.1. The work done against friction is??

please help, thanks!

Answer 1

First, solve for the force of friction using the formula,

f=uR

where u is the coefficient of friction, R the force normal to the plane, and f is the force of friction.

Substitute known values:

f=0.1*6*g*cos30
=0.1*50.92
= 5.09N

Then solve for the work done using the formula,

Work=f*s

where s is the displacement

Substitute known values:

Work=5.09*8=40.7J

By the way, if you are a student, and this problem was given by your teacher, please tell him or her that there's something not right about how the problem is worded.

Remember that for the box to slide "at a constant speed" the force applied should be equal to the force of friction. In this case, the force applied is 50N while the force of friction is only 5.09N. Thus the box will accelerate upwards along the incline. A body that accelerates CANNOT be said to be moving AT A CONSTANT SPEED.

So how do we word the problem correctly? Just delete the words "at a constant speed."

Answer 2

The friction force Ff= 0.1 * g * 6kg * cos(30deg).
The work done against friction is 8m * Ff.
Extra info: Ff is less than the applied force. The extra force acts against the gravitational force along the ramp (g * 6kg * sin(30deg)) and accelerates the box.