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Power/Work problem with some Newton's Second Law

  • Thread starter StormDuck
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  • #1
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Homework Statement


A 6.7-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.
(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?

Correct answer: 0.131 kW

(b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.6 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?

Correct answer: 0.713 kW


Homework Equations



W=|F||d|cos(θ)
Power = Work/Time
∑F = ma


The Attempt at a Solution



For a, I've solved correctly;
P=T*v = Tvcos(Θ)
P=T*v (cos(0°) = 1, so I removed it)

∑F = ma
a = 0
∑F = T - W = T- (m*g) = 0
∑F = T - (6.7kg*9.81 m/s^2) = 0
T = 65.727 N
P = 65.727 N * 2.0 m/s = 131.454 W = .131454 kW

for b, I'm doing something wrong;

I started with using the a kinematic equation to solve for a

s = V(o)t + .5at^2

1.6m/(.5*(0.42s)^2) = a = 18.14058957 m/s^2

F = ma
F = 6.7kg(18.14058957 m/s^2) = 121.5419501 N
Work = |F||d|
W = 121.5419501 N * 1.6 m = 194.4671202 J
P = Work/time
P = 194.4671202 J / .42 s = 463.0169528 W = .4630169528 kW

Which is obviously wrong from the correct answer that is given....any help would be greatly appreciated. Thanks in advance.
 
Last edited:

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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475

Homework Statement


A 6.7-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.
(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?

Correct answer: 0.131 kW

(b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.6 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?

Correct answer: 0.713 kW


Homework Equations



W=|F||d|cos(θ)
Power = Work/Time
∑F = ma


The Attempt at a Solution



For a, I've solved correctly;
P=T*v = Tvcos(Θ)
P=T*v (cos(0°) = 1, so I removed it)

∑F = ma
a = 0
∑F = T - W = T- (m*g) = 0
∑F = T - (6.7kg*9.81 m/s^2) = 0
T = 65.727 N
P = 65.727 N * 2.0 m/s = 131.454 W = .131454 kW

for b, I'm doing something wrong;

I started with using the a kinematic equation to solve for a

s = V(o)t + .5at^2

1.6m/(.5*(0.42s)^2) = a = 18.14058957 m/s^2

F = ma
F = 6.7kg(18.14058957 m/s^2) = 121.5419501 N
Work = |F||d|
W = 121.5419501 N * 1.6 m = 194.4671202 J
P = Work/time
P = 194.4671202 J / .42 s = 463.0169528 W = .4630169528 kW

Which is obviously wrong from the correct answer that is given....any help would be greatly appreciated. Thanks in advance.
In part b when using newton 2, you have calculated the net force acting on the box. But the problem is looking power delivered by the man's force.
 
  • #3
6
0
In part b when using newton 2, you have calculated the net force acting on the box. But the problem is looking power delivered by the man's force.
Thanks a ton, that was all I needed, coming up with the right answer now.
 

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