Work and Power of pulling student

[Ki-nana18]

Homework Statement

Using a force of 12.0 Newtons, a student pulls a 50.0 Newton weight along the table top for a distance of 10.0 meters in 4.0 seconds. Compute the power developed by the student.

Homework Equations

W=Fcos(theta)d
P=W/t

The Attempt at a Solution

W=(12.0)(cos(50))(10)
W=115.8 J

P=115.8/4.0
P=28.95 J/s

 

[Doc Al]

W=(12.0)(cos(50))(10)

Where did you get the angle = 50°?

 

[Ki-nana18]

I thought since the weight is 50, that was the angle. So it should just be (12.0 N)(10.0 m) which equals 120. J? Then the power will be 120.J/4.0s which will equal 30. J/s or watts, I presume?

 

[Doc Al]

I thought since the weight is 50, that was the angle.

No, you cannot draw any such conclusion. The angle in question is that between the force and the displacement. Assuming the force is exerted in the direction of the displacement (parallel to the table top) then the angle will be 0.

So it should just be (12.0 N)(10.0 m) which equals 120. J? Then the power will be 120.J/4.0s which will equal 30. J/s or watts, I presume?

Right, assuming an angle of 0° as discussed above.

 

[jegues]

Like Doc Al had pointed out, your confusion is with the concept of weight.

Weight is simply the gravitational force applied to an object with a given mass.

If I told you I had an object whos weight was 29.4N then it's mass would simply be 3kg.

Can you see how I was able to solve for the mass?

You must have a clear understanding of the difference between weight and mass and the properties they contain before moving on so hammer this stuff into your head!

 

[Ki-nana18]

I'm sorry, but I don't understand how you were able to solve for mass. How did you do that?

 

[Doc Al]

Weight = mass*g

Given the weight of an object, you can solve for its mass and vice versa.

 

[Ki-nana18]

Thank you for all clarification!