# Relating Power to Work and KE...

by glockstock
Tags: power, relating, work
 P: 5 1. The problem statement, all variables and given/known data A 6.1-kg box is being lifted by means of a light rope that is threaded through a single, light, frictionless pulley that is attached to the ceiling. (a) If the box is being lifted at a constant speed of 2.0 m/s, what is the power delivered by the person pulling on the rope? (b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.4 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope? 2. Relevant equations P = F * v KE = .5mv^2 W = F * distance W = KE 3. The attempt at a solution I've answered the first problem, getting .120 kW as an answer. But I'm having trouble with the (b), and maybe the concept of average power is throwing me off. I don't know what to do with the time given (0.42 s) and the height (1.4 m)...I know power is KE/T, but I don't know what the velocity is in KE since no acceleration was given. Thanks for your help! Much appreciated.
 HW Helper PF Gold P: 3,440 If the box is lifted at constant acceleration from rest, you can find the acceleration using the kinematic equation for x(t).
 P: 5 OK, so using the kinematic equation I've found the acceleration to be 15.87 m/s^2....do I use another kinematic equation to find a velocity, and use F=ma with the mass of the box (6.1kg) and my acceleration (15.87) to get a force of 96.81N, then multiply that by the velocity for power? Or do I have to consider the force of gravity too, and subtract that force from the 96.81N...
 HW Helper PF Gold P: 3,440 Relating Power to Work and KE... You need to consider the force of gravity. F = ma gives you the net force not the force the person is exerting on the box. Once you find the force exerted by the person, you need to multiply by the average velocity to get average power.
 P: 4 P=Fv=ma(vavg) SOLVE FOR a:h=$$\frac{1}{2}$$at2PLUG IN a and SOLVE FOR vavg (which is v/2):v2=2ah
HW Helper
PF Gold
P: 3,440
 Quote by Rubber Band P=Fv=ma(vavg) SOLVE FOR a:h=$$\frac{1}{2}$$at2PLUG IN a and SOLVE FOR vavg (which is v/2):v2=2ah
The power that Rubber Band proposes as the answer is not the power delivered by the person. It is the power delivered by the net force. The person has to deliver enough power to do two things, overcome gravity and accelerate the box. The tension in the rope is equal and opposite to the force the person exerts. Therefore, the power delivered by the rope is the same as the tension. When a rope accelerates a bucket up (or down), the tension is not ma.

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