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Lifting a Box - Work and Power

by cowmoo32
Tags: lifting, power, work
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cowmoo32
#1
Oct17-12, 07:10 PM
P: 120
1. The problem statement, all variables and given/known data
A 7-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. If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.5 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?


2. Relevant equations
(1) d = V0t + 0.5at2
(2) F=ma
(3) W = Fd
(4) P = W/Δt

3. The attempt at a solution
(1) a = 17m/sec2

(2)
F = 7*17
F = 119

(3)
W = 119*1.5
W = 178.5

(4)
P = 178.5/0.42
P = 425

Here's the thing, that answer agrees with the equation in the solution manual, but not on my homework. I'm reworking an old problem and I got it correct the first time around and can't figure out how I arrived at the answer, which is 0.67kW.
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cepheid
#2
Oct17-12, 08:46 PM
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P: 5,197
Quote Quote by cowmoo32 View Post
1. The problem statement, all variables and given/known data
A 7-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. If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.5 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?


2. Relevant equations
(1) d = V0t + 0.5at2
(2) F=ma
(3) W = Fd
(4) P = W/Δt

3. The attempt at a solution
(1) a = 17m/sec2

(2)
F = 7*17
F = 119

(3)
W = 119*1.5
W = 178.5

(4)
P = 178.5/0.42
P = 425

Here's the thing, that answer agrees with the equation in the solution manual, but not on my homework. I'm reworking an old problem and I got it correct the first time around and can't figure out how I arrived at the answer, which is 0.67kW.
Method 1: Using Forces

119 N is the NET force that is applied to the object. Therefore the work you calculated is the NET work done on the object. However, in order to get this net force upward, the person has to apply this force PLUS the weight of the object. So he does more work than what you computed, because he or she also has to do work against gravity. The work done against gravity goes into increasing the object's potential energy, whereas the net work done by the net force goes into increasing its kinetic energy. Anyways:

Fnet = Fapplied + weight = Fapplied - mg

Fapplied = Fnet + mg = 119 N + (7 kg * 9.81 N/kg)

= 187.67 N

Wapplied = (187.67 N * 1.5 m) = 281.5 J

P = W/Δt = 281.5 N/0.42 s = 670.25 W

Method 2: Using Energies

This method is not really different mathematically, it's only different conceptually. The power is the rate at which energy is delivered by the person, so to find the average power, just take the total energy expended by the person, and divide it by the time interval. The total energy is the amount by which the energy of the object increases, which is the sum of its changes in potential and kinetic energies:

E = (1/2)mvf2 + mgh

vf2 = 2ad = 2*(17 m/s2)*(1.5 m) = 51 m2/s2

E = 0.5*(7 kg)*(51 m2/s2) + (7 kg)*(9.81 N/kg)*(1.5 m) = 281.5 J.

P = E/Δt = 670.25 W
cowmoo32
#3
Oct18-12, 11:15 AM
P: 120
Thanks!


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