Incline plane and friction prob.?

Question

A 6.0 kg box is sliding up a 8.0 m long friction incline at a constant speed at an angle of 30 degrees by a 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

Well, you need to find the Normal force :

the weight of the box is : 6*9.8 m/s^2 = 58.8 Newtons

but that's not the normal force, if you make a graphic, you will notice, that the normal force will be :

N = 58.8*cos(30) = 50.9 Newton

But there is a force of : 50 Newton, parallel to the plane :

This force, is parallel, the normal force, is perpendicular to the block, so this force, doesn't affect the normal force, make the graphic, and you will notice it.

The force of friction = 50.9 * 0.1 = 5.09 Newtons

The box is sliding at a constant speed, but the problem is asking for the work done by the force of friction :

W = 5.09*8 = 40.72 Joules

Hope that might help you

Answer 2

Over the 8.0 m distance, Wf = f * s, where s = 8.0 m.
To find f, we need the component of the box's weight perpendicular to the incline. It's weight, w, is m * g, or 8.0 x 9.8 kg*m/s^2.
That is, w = 8.0 x 9.8 N.
Now resolve this into components perpendicular (and parallel) to the incline:
wperp = w * cos 30. We don't need the parallel component.
Then, the friction force f = wperp x 0.1, or

f = w*cos(30) x (0.1), and
the work due to friction is f x s = w*cos(30) x (0.1) x 8 N-m, that is, Newton-meters.

Answer 3

if 50N force is acting \\ to slope then 50sin(30) is actinng perpendicular to box and also mgcos(30) due to force of gravitation
50sin(30)+6*10*1/2=50*1/2+30=55N
u=f/r
u=f/55N
f=55*.1
f=5.5 N
w=f*s
w=8*5.5=44N