Work Done: Which Initial Velocity is Most Effective?

[vivekfan]

Homework Statement

1) In which of the following situations is work done?

A man rubs his hands together and they get warm or the moon orbits the earth?

2) A man pulls a 100 kg block across a frictionless floor using a rope. The rope makes an angle of 60 degrees with the horizontal and the man pulls with a force of 50 N. The block moves across the surface 20 m.

Assume the force acts for a time t. Assume the block has initial velocity v. Initial velocity in which direction would result in the greatest amount of work done on the block?

a) Initial velocity to the left
b)Initial velocity to the rightt
c)Initial velocity in either direction would result in the same amount of work done because distance traveled would be the same.
d)Initial velocity in either direction would result in the same amount of work done because the time t is constant.
e)none of the above.

Homework Equations

W=Fdcos(theta)

The Attempt at a Solution

For the first question, I'm pretty sure the only work done is by the man rubbing his hands together, because it's work done by friction. The moon orbiting the Earth is a circular path, so the displacement would be tangent (or perpendicular) to the inward force, right? I'm not sure if my reasoning is correct.

For the second question, I'm pretty confused.

I don't understand what velocity has to do with work, given the equation, and I'm not sure what's really happening in the question with regard to forces. Please help. Thanks!

 

[quantumdude]

The Attempt at a Solution

For the first question, I'm pretty sure the only work done is by the man rubbing his hands together, because it's work done by friction. The moon orbiting the Earth is a circular path, so the displacement would be tangent (or perpendicular) to the inward force, right? I'm not sure if my reasoning is correct.

Yes, that's right.

For the second question, I'm pretty confused.

I don't understand what velocity has to do with work, given the equation, and I'm not sure what's really happening in the question with regard to forces.

Work and velocity are related via the work-energy theorem. The work [itex]W[/itex] done on the box is equal to the change in kinetic energy [itex]\Delta K[/itex] of the box. So...


[tex]W=\Delta K[/tex]

[tex]W=K_f-K_i[/tex]

[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/tex]

 

[vivekfan]

Yes, that's right.



Work and velocity are related via the work-energy theorem. The work [itex]W[/itex] done on the box is equal to the change in kinetic energy [itex]\Delta K[/itex] of the box. So...


[tex]W=\Delta K[/tex]

[tex]W=K_f-K_i[/tex]

[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/tex]

But what does this have to do with the direction of initial velocity?