How Fast Must a Student Climb to Match a Lightbulb's Power Output?

[Knfoster]

Homework Statement

A 54.2 kg student climbs a 6.95 m rope and stops at the top. What must her average speede in order to match the power output of a 110 W lightbulb?

Homework Equations

PE=mgh
KE=1/2mv^2
W=(Xf-Xi)(cos theta)(F)
W= delta KE
PE+KE = PE+KE (before and after)

The Attempt at a Solution

I know that the PE for the student is 3691.6 J at the top of the rope. And KE is zero at the top of the rope, so PE at the bottom of the rope would be 0, and KE would be 3691.6.

I don't know what to do next and I don't know how to use the 110 W light bulb.

Please help. Thank you.

 

[Doc Al]

I know that the PE for the student is 3691.6 J at the top of the rope.

Good. This represents the work she must do to raise herself up the rope. Hint: How quickly must she climb the rope so that her power equals 110 W? (Find the time, then figure out her speed.)

 

[Knfoster]

How do I go about finding the time? Is there a certain equation that I need?

 

[Knfoster]

Do I use one of the four equations for motion?

 

[Doc Al]

How do I go about finding the time? Is there a certain equation that I need?

What's the definition of power?