Work, Energy and Power (Work Problem)

[matadorqk]

Homework Statement

A man drags a sack of flour of mass 50kg at a constant speed up an inclined plane to a height of 5.0m. The plane makes an angle of 30 degrees with the horizontal, and has a constant frictonal force of 200N, which acts on the sack down the plane. Calculate the work the man has dne against the friction, and against gravity.

Homework Equations

W=(F)(S) (Where f=force and s=displacement)
W=(F\cos\vartheta)(S)
**Also usage of SOHCAHTOA to solve the triangle.

The Attempt at a Solution

Well, first I drew a triangle for the problem, and obtained that he travels a total of 10m going upward 30 degrees to the horizontal.
Therefore, I solved it like this:
Against friction, the force is 200N, so (200cos30)(10) = 1732J.
Against gravity, the force is (50)(9.8) so (490cos30)(10)=4244J.

**I am unclear overall if I am doing this right, some help?

 

[matadorqk]

As I further tried solving, wouldn't the work done against gravity be the following:

W_{g}=mgh=(50)(9.8)(5)=2450J

? Any ideas?

 

[PhanthomJay]

As I further tried solving, wouldn't the work done against gravity be the following:

W_{g}=mgh=(50)(9.8)(5)=2450J

? Any ideas?

Yep, that's correct, nice job. Now, calcualte the work done against friction. HINT: How much work is done by the friction force?

 

[matadorqk]

Yep, that's correct, nice job. Now, calcualte the work done against friction. HINT: How much work is done by the friction force?

The friction force does a force of 200N. So, if we plot that into our equation for work, its
W=200s, but its an inclined plane, so the W=(200cos30)(s), where s=10, so when I solve for work, I get that is 1732J, is that right?

 

[PhanthomJay]

The friction force does a force of 200N. So, if we plot that into our equation for work, its
W=200s, but its an inclined plane, so the W=(200cos30)(s), where s=10, so when I solve for work, I get that is 1732J, is that right?

No. You are correct that s=10. You are also correct that the friction force is 200N. And you are also corect tnat the work done by the friction force is 200s.. But then you incorrectly throw cos 30 in there. Remember that work is force times distance in the direction of the force.

 

[matadorqk]

No. You are correct that s=10. You are also correct that the friction force is 200N. And you are also corect tnat the work done by the friction force is 200s.. But then you incorrectly throw cos 30 in there. Remember that work is force times distance in the direction of the force.

Ohh, so the frictional force is solely pushing downwards on the sack, therefore meaning that the frictional force is an upward force, so the frictional force of the object would be 2000J?

 

[PhanthomJay]

Ohh, so the frictional force is solely pushing downwards on the sack, therefore meaning that the frictional force is an upward force, so the frictional force of the object would be 2000J?

Oh, now your mixing up your directions. Yes the friction force is down the plane. It is not an upward force, it is a downward force. So the man is exerting an upward force to counteract it (and gravity). 2000J is correct for the work done by the man against friction. So what is the total work done by the man against friction and gravity??

 

[matadorqk]

Oh, now your mixing up your directions. Yes the friction force is down the plane. It is not an upward force, it is a downward force. So the man is exerting an upward force to counteract it (and gravity). 2000J is correct for the work done by the man against friction. So what is the total work done by the man against friction and gravity??

Well, he does 2000J against friction, and 2450J against gravity, for a total work of 4450J.

 

[PhanthomJay]

Well, he does 2000J against friction, and 2450J against gravity, for a total work of 4450J.

Yup, that'll do it.

 

[matadorqk]

Yup, that'll do it.

Thanks so much for your help.