Calculating Crane Power with Constant Velocity: Two Homework Problems

[alex7298]

Homework Statement

2 problems.
1: A man lifts a 91kg person straight up a distance of 1.2m in 0.51s. What power is used to do this?

2: 3.00x10^2kg piano is lifted at a steady velocity from ground to a height of 10.0m. The crane doing the lifting has a steady power of 4.00x10^2W. What time does it take to do this?

Homework Equations

Both equations should be able to use P=Fd/t

The Attempt at a Solution

With both problems, the book states constant velocity so the F=ma=0, which doesn't make sense because that would make the answer 0.
Is there another way of finding force that i am forgetting?
Thanks

 

[PhanthomJay]

Homework Statement

2 problems.
1: A man lifts a 91kg person straight up a distance of 1.2m in 0.51s. What power is used to do this?

2: 3.00x10^2kg piano is lifted at a steady velocity from ground to a height of 10.0m. The crane doing the lifting has a steady power of 4.00x10^2W. What time does it take to do this?

Homework Equations

Both equations should be able to use P=Fd/t

The Attempt at a Solution

With both problems, the book states constant velocity so the F=ma=0, which doesn't make sense because that would make the answer 0.
Is there another way of finding force that i am forgetting?
Thanks

That makes the net force and the net work = 0. But you're not looking for the net, you're looking for the work done by just one of the forces against gravity.

 

[alex7298]

would I use GPE=mgh for the work?
so for problem 1: work=(91kg)(9.8)(1.2)=1070.16J===> P=1070.167J/0.51s=2098.366W?

 

[PhanthomJay]

would I use GPE=mgh for the work?
so for problem 1: work=(91kg)(9.8)(1.2)=1070.16J===> P=1070.167J/0.51s=2098.366W?

Yes, correct, as long as you understand why you are equating the work done by the man to the potential energy change of the person.