Ratio of work done using pulley

[negation]

Homework Statement

As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box.

The Attempt at a Solution

a) Once you have pulled hard enough to start the box moving upward, what is the magnitude F of the upward force you must apply to the rope to start raising the box with constant velocity?
Express the magnitude of the force in terms of m, the mass of the box.

F = 4.9 m?

b) Part B

Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a pulley (as in the previous part).
What is Wd/Wp, the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?
Express the ratio numerically.

I know that in both cases, the work done by lifting the box directly or by using the pulley are equivalent since in lifting the box, the work done is wd = Fxd = mgh.
In using a pulley, the force is halved but the distance is 2h.

 

[haruspex]

F = 4.9 m?

Since no units have been specified, you cannot plug in g = 9.8. Just leave it as g.

b) Part B

Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a pulley (as in the previous part).
What is Wd/Wp, the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?
Express the ratio numerically.

I know that in both cases, the work done by lifting the box directly or by using the pulley are equivalent since in lifting the box, the work done is wd = Fxd = mgh.
In using a pulley, the force is halved but the distance is 2h.

Right, so your answer is?

 

[negation]

Since no units have been specified, you cannot plug in g = 9.8. Just leave it as g.

Right, so your answer is?

In response to part(a): would mg/2 do fine?

lifting directly:

w = fxh = mgh

using pulley: F + T = mg
2F = mg
F = mg/2
but distance is now 2h
2h(mg/2) = mgh
w = mgh
Since work done are both the same, then the ratio has to be 1?

 

[haruspex]

In response to part(a): would mg/2 do fine?

lifting directly:

w = fxh = mgh

using pulley: F + T = mg
2F = mg
F = mg/2
but distance is now 2h
2h(mg/2) = mgh
w = mgh



Since work done are both the same, then the ratio has to be 1?

Yes, and yes.