# KE, average speed

"A 48.5 kg student climbs a 5.90 m long rope and stops at the top. What must her average speed be in order to match the power output of a 140 W lightbulb?"

Here's how I approached this problem:

Since 1 W = 1 J/s, I set KE = 140 J.

Then, v^2 = KE/((.5)m) = 140/24.25 = 5.77

Then v = 2.4 m/s

But that's not the right answer.

What am I doing wrong?

Â© BrainMass Inc. brainmass.com March 4, 2021, 5:46 pm ad1c9bdddfhttps://brainmass.com/physics/energy/ke-average-speed-9947

#### Solution Preview

Hi there,

<br>It looks like you're trying to use the work-kinetic energy theorem to solve this problem (i.e. that work done = change in kinetic energy) or something similar. But actually, power is the RATE OF CHANGE of work being done, or equivalently, the rate of change of energy. And if you just let someone move with a constant speed, they're ...

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