Work problem

[tigerseye]

An 8.0 kg crate is pulled up a 30 degree incline by a person pulling on a rope that makes an 18 degree angle with the incline. The tension in the rope is 120 N and the crate's coefficient of kinetic friction on the incline is 0.25. If the crate moves 5.0 m, what is the work done by a.) friction and b.) the normal force?

I calculated the work done by tension in the rope to be 571 J and the work done by gravity to be -196 J. How do I find the friction and normal force?

[bobp718]

If you have set things up nicely (namely having the normal force perpendicular to the inclined plane surface) the work done by the normal force should be zero because

\overrightarrow{F}\centerdot\overrightarrow{r}=W

this reduces to

F\times r \times\cos(\theta)=W

because the angle that the normal force makes with the position vector (or the time derivative of it anyway) is 90^\circ the work is zero.

As for the work done by friction that will simply be

W_{friction}=\mu_{k}(F_{A}\sin(\theta_{1})-mg\sin(\theta_{2}))\Delta s

where \theta_{1} is the angle that the force is being applied parallel to the plane and \theta_{2} is the angle of the plane parallel to horizontal.

[marlon]

Suppose the x-axis is along the incline and y-axis is perpendicular to this surface...


We have 4 fources on the object : 120N, gravity, friction{\mu}{\vec N}, normal force N

So along the x-axis we have :
F_x = -mgsin(\theta) - {\mu}N + 120cos(18)

Along the y-axis we have :
F_y = 0 = -mgcos(\theta) + N+120sin(18)

What you need to know is the magnitude of the friction force, which is equal to {\mu}N = {\mu}mgcos(\theta).


regards
marlon

[learningphysics]

Suppose the x-axis is along the incline and y-axis is perpendicular to this surface...


We have 4 fources on the object : 120N, gravity, friction{\mu}{\vec N}, normal force N

So along the x-axis we have :
F_x = -mgsin(\theta) - {\mu}N + 120

Along the y-axis we have :
F_y = 0 = -mgcos(\theta) + N

What you need to know is the magnitude of the friction force, which is equal to {\mu}N = {\mu}mgcos(\theta).


regards
marlon

Note that the 120N force is making an angle of 18 degrees with the inline. So the equations should be:

F_x = -mgsin(\theta_1) - {\mu}N + 120cos(\theta_2)

F_y = 0 = -mgcos(\theta_1) + N + 120sin(\theta_2)

where \theta_1=30, and \theta_2=18

From the F_y equation we get:
N=mgcos(\theta_1)-120sin(\theta_2)
Then you get the frictional force {\mu}N

[marlon]

thanks for the correction learningphysics...

marlon