Work by Gravity and Tension forces on an Incline

[vtstudent]

1. An 8.3 kg crate is pulled 4.8 m up a 30 degree incline by a rope angled 18 degrees above the incline. The tension in the rope is 120 N and the crate's coefficient of kinetic friction on the incline is 0.24. How much work is done by tension, by gravity, and by the normal force? What is the increase in thermal energy of the crate and incline?



2. W=F$$_{}x$$d$$_{}s$$
F$$_{}g$$=mg
F$$_{}fr$$=uN
F$$_{}N$$=mgsin30

The Attempt at a Solution

I think the work done by the normal force is 0 because it is perpendicular to the motion of the crate.

For the work done by gravity, I found mg = 8.3*9.8=81.34
Then, for the component of gravity parallel to the motion of the crate, I found 81.34sin30=40.67. I multiplied this by distance to get 40.67*4.8=195.216 as the work done by gravity

For work done by the tension force, I found the component of tension parallel to the motion of the crate. 120cos18=114.126
Then, I multiplied this by distance to get 547.81 as the work done by the tension force.

Now, to deal with friction I am a bit lost. I found the normal force. Normal force = mgcos30 = 8.3(9.8)cos30=70.44
Then I found the friction force. F(friction)=.24(70.44)=16.91N
Do I multiply this by distance (16.91*4.8=81.15N) and then add it to the work done by the tension force, because the tension force must overcome the friction force as well?

 

[collinsmark]

Hello vtstudent,

Welcome to Physics Forums!

I think the work done by the normal force is 0 because it is perpendicular to the motion of the crate.

Very nice!

For the work done by gravity, I found mg = 8.3*9.8=81.34
Then, for the component of gravity parallel to the motion of the crate, I found 81.34sin30=40.67. I multiplied this by distance to get 40.67*4.8=195.216 as the work done by gravity

Almost. Very close. But don't forget about the direction. Ask yourself this: Is the gravitational force in the same direction as the y-component of displacement or is it in the opposite direction?

Another way of asking the same question, is the component of gravitational force that is parallel to the displacement 81.34sin30 or is it 81.34sin(-30)?

Still another way of asking the same question, is gravity doing actual work on the crate or is it the other way around?

For work done by the tension force, I found the component of tension parallel to the motion of the crate. 120cos18=114.126
Then, I multiplied this by distance to get 547.81 as the work done by the tension force.

Sounds good to me.

Now, to deal with friction I am a bit lost. I found the normal force. Normal force = mgcos30 = 8.3(9.8)cos30=70.44

That is not the normal force. Remember, the rope is pulling the crate away from the surface a little bit. You must take that into account too.

Then I found the friction force. F(friction)=.24(70.44)=16.91N
Do I multiply this by distance (16.91*4.8=81.15N) and then add it to the work done by the tension force, because the tension force must overcome the friction force as well?

No, once you find the friction force just multiply it times the displacement (or the component of displacement parallel to the friction force -- which in this case is easy since they are parallel to begin with). That gives you the work done by the friction force itself.

 

[vtstudent]

Almost. Very close. But don't forget about the direction. Ask yourself this: Is the gravitational force in the same direction as the y-component of displacement or is it in the opposite direction?

Another way of asking the same question, is the component of gravitational force that is parallel to the displacement 81.34sin30 or is it 81.34sin(-30)?

So the gravitational force is negative 81.34sin30. My problem is that this is for Mastering physics and I need to put all three answers in the same answer box. Then when I get it wrong, I need to figure out which part is wrong! such a poor design!


That is not the normal force. Remember, the rope is pulling the crate away from the surface a little bit. You must take that into account too.

I don't know how to calculate normal force any other way - I don't think we've gone over this in class (although my class is consistently more simplistic than the homework...)

No, once you find the friction force just multiply it times the displacement (or the component of displacement parallel to the friction force -- which in this case is easy since they are parallel to begin with). That gives you the work done by the friction force itself.[/QUOTE]

Thank you so much for this help!

 

[collinsmark]

I don't know how to calculate normal force any other way - I don't think we've gone over this in class (although my class is consistently more simplistic than the homework...)

You can do it!

But it's important that you draw a free body diagram. Take a look at all the forces acting on the crate. Then sum the force components which are perpendicular to the surface.

mgcosθ is part of the normal force. This term comes from the gravitational force mg. The component of mg that is perpendicular to the surface is mgcosθ. Now do the same thing with the tension force from the rope. What is the component of the tension that is perpendicular to the surface? Once you have that, sum the two force components together (the one from gravity and the one from the rope tension). Be careful about signs, by the way -- one of the components is directed into the surface and the other away from it.

 

[rwisz]

For the increase in thermal energy, I think we can use the formula Q=mcdeltaT. We would need to find the specific heat capacity and the change in temperature, but I am not sure how to go about finding these values. It would also depend on the material of the crate and incline. Additionally, some of the energy may be lost as heat due to friction, so the increase in thermal energy may not be equal to the total work done.