How Much Work Is Done on a Lift Pulled Up by a Cable?

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Homework Help Overview

The discussion revolves around a physics problem involving a lift being pulled upwards by a cable, with a block placed inside it. The problem includes parameters such as the masses of the block and the lift, the distance the lift is pulled, and the normal force acting on the block. Participants are exploring the work done on the lift and the implications of the forces involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate the work done using the normal force and gravitational force, while questioning the relationship between work, potential energy, and kinetic energy. There are discussions about the implications of the forces acting on the block and the lift, and how these relate to acceleration and speed.

Discussion Status

The discussion is active, with various approaches being explored. Some participants are providing insights into the relationship between forces and energy, while others are questioning the assumptions made in the problem. There is a recognition of the complexity of the situation, particularly regarding the effects of gravity and acceleration.

Contextual Notes

Participants note the absence of initial speed in the problem statement, which raises questions about how to approach the calculation of kinetic energy. The discussion also highlights the need to consider both gravitational and applied forces when analyzing the work done on the lift.

Full Beige
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Homework Statement


a block of mass m = 20 kg is placed on the floor of a lift (mass M = 1000 kg), the lift is being pulled up distance d =25m by a cable, the normal force on the block FN has a constant magnitude of 300N, how much work is done on the lift by the force on the cable?

Homework Equations



Wgravity = -mgd m=mass, d = distance, g= 9.8

W = Fd

Wnet = \sum Work

The Attempt at a Solution



For Normal force I used W = Fd >> 300N x 25m = 7500J

For Wgravity I did >> -(20+1000) x 9.8 x 25 = -249900 J

Then Wnet = -Wgravity + W = -249900 + 7500 = -242400 J or -242.4 KJ

for some reason I don't think this is right? can anybody give me any advice please? I have not been provided with the actual answers.
 
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Careful.

The work apparently resulted in an increase in potential energy all right. But it also apparently resulted in an increase in kinetic energy as well.

At 25m what is the speed of the elevator?
 
LowlyPion said:
Careful.

The work apparently resulted in an increase in potential energy all right. But it also apparently resulted in an increase in kinetic energy as well.

At 25m what is the speed of the elevator?

lowlypion there is no speed given in the question. if I work it out then would it be V = \sqrt{2K / m} , what value would I use for Kinetic energy? Wnet or W?
 
Hi, since LowlyPion is offline, I'll give You a clue:
the lift is being pulled up from initial velocity 0, now it had to have speed after X meters,
how can You find acceleration with a mass of a body and the normal force exerted by the floor on the body?and than you should find the velocity.
good luck
 
Full Beige said:
lowlypion there is no speed given in the question. if I work it out then would it be V = \sqrt{2K / m} , what value would I use for Kinetic energy? Wnet or W?

Actually you are right, in that they don't give the initial speed and so that is not a good way to go. My comment was really directed at the notion that the work against gravity not relating to the total energy should be a clue that work was going into increasing kinetic energy.

So look at it another way.

If the force on the block is 300N then the force on the cable will be in the same ratio.

Hence I think you can say

300N / 20 kg = Fc / (1000 + 20)

Fc = 15*(1020)

So work by the cable over the same distance should look like

W = Fc * d
 
LowlyPion said:
Actually you are right, in that they don't give the initial speed and so that is not a good way to go. My comment was really directed at the notion that the work against gravity not relating to the total energy should be a clue that work was going into increasing kinetic energy.

So look at it another way.

If the force on the block is 300N then the force on the cable will be in the same ratio.

Hence I think you can say

300N / 20 kg = Fc / (1000 + 20)

Fc = 15*(1020)

So work by the cable over the same distance should look like

W = Fc * d

Just wanted to approve the answer.
done in a different way and got the same answer.
If you're interested:
m=20kg weigh in non accelerating lift = 200 N , in this one 300N thus acceleration = 5m/s^2
using kinematics V^2=250
W= change in Ke and Ep= m*250/2 + m*250. while m=1020.
same answer! awesome :D
IMO LowlyPion's method is better, although i didn't quite understood it...
 
Dweirdo said:
Just wanted to approve the answer.
done in a different way and got the same answer.
If you're interested:
m=20kg weigh in non accelerating lift = 200 N , in this one 300N thus acceleration = 5m/s^2
using kinematics V^2=250
W= change in Ke and Ep= m*250/2 + m*250. while m=1020.
same answer! awesome :D
IMO LowlyPion's method is better, although i didn't quite understood it...

Sure you do.

F = m*a

Acceleration is (must be) the same.
 
LowlyPion said:
Sure you do.

F = m*a

Acceleration is (must be) the same.

awww , awesome ! stupid me XD
thanks
 
LowlyPion said:
Actually you are right, in that they don't give the initial speed and so that is not a good way to go. My comment was really directed at the notion that the work against gravity not relating to the total energy should be a clue that work was going into increasing kinetic energy.

So look at it another way.

If the force on the block is 300N then the force on the cable will be in the same ratio.

Hence I think you can say

300N / 20 kg = Fc / (1000 + 20)

Fc = 15*(1020)

So work by the cable over the same distance should look like

W = Fc * d

hmm I can see what you are saying that the two forces can be the same, but I don't quite understand why the Work gravity does not affect it? since the lift is moving upwards, does gravity not affect it? sorry if its a stupid question. thanks two both of you guys for your help.
 
  • #10
Full Beige said:
hmm I can see what you are saying that the two forces can be the same, but I don't quite understand why the Work gravity does not affect it? since the lift is moving upwards, does gravity not affect it? sorry if its a stupid question. thanks two both of you guys for your help.

Of course gravity enters into it. And some of the work of the lift goes into raising the mass over the 25m. But the statement of the problem indicates that there is an excess of energy developed, that will go into kinetic energy.

Looking at the weight of the block on the floor you can see that the 20 kg times 9.8 is short of the 300N force given from the motion. So part of the 300N is gravity induced (196N of it ) and the remainder ( 104N ) goes into acceleration of the block.

Well if the block is accelerating then so is the lift. And if the lift is accelerating then the lift's mass acceleration plus its weight at rest, will be in the same proportion as the blocks acceleration and its weight at rest.
 
  • #11
LowlyPion said:
Of course gravity enters into it. And some of the work of the lift goes into raising the mass over the 25m. But the statement of the problem indicates that there is an excess of energy developed, that will go into kinetic energy.

Looking at the weight of the block on the floor you can see that the 20 kg times 9.8 is short of the 300N force given from the motion. So part of the 300N is gravity induced (196N of it ) and the remainder ( 104N ) goes into acceleration of the block.

Well if the block is accelerating then so is the lift. And if the lift is accelerating then the lift's mass acceleration plus its weight at rest, will be in the same proportion as the blocks acceleration and its weight at rest.

Omg I never saw it like that, how'd you get so smart lol. Thanks a lot lowlypion.
 

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