Finding Force and work on a moving object

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The discussion centers around calculating the force, displacement, and work done on a 5000 kg elevator being lifted at a constant velocity of 2.0 m/s. The correct force exerted by the cable is 49,000 N, which balances the gravitational force, resulting in a net force of zero. The displacement of the elevator after 15 seconds is determined to be 30 meters. The work done by the cable against gravity is calculated to be approximately 1.47 million joules. The calculations highlight the importance of distinguishing between speed and acceleration in physics problems.
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Homework Statement



A cable lifts a 5000kg elevator at a constant velocity of 2.0m/s [up]

I) Find the force acting on the elevator.
II) Find displacement of the elevator in 15s.
III) find the work done by the cable on the elevator over a time period of 15s.

Homework Equations


W=FΔd
Δd=ta

The Attempt at a Solution


I)
F=mg+ma
F=(5000*9.8)+(5000*2)
F=59'000 N
II)
30 m up
III)
W=FΔd
W=(59,000)(30)
W=1.77 million joules

*updated is this correct now?*
 
Last edited:
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No.

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Unfortunately there is no simple yes or no answer because the question is ambiguous. Does "the force" mean the force from the cable or the net of all forces?
 
haruspex said:
Unfortunately there is no simple yes or no answer because the question is ambiguous. Does "the force" mean the force from the cable or the net of all forces?

Alright so i will guess that "the force" means the force from the cable so...
F=mg
F=5000 x 9.8m/s
F=49'000 N

Is that correct? or am i missing something with the 2m/s the elevator is moving. Thanks for the help.
 
Hello!
The elevator is moving with constant velocity, so the net force acting on it is 0. (Ans 1.)
Also, work done by the cable equals the work against gravity (forces are equal!). (Ans.3)

i)The force is $$|mg|=|F_{upward}|$$

ii) 15s * 2m/s = 30m

iii) $$mgh = mg * 30m = 5000kg * 9.8m/s^2 * 30m = 1.47*10^6 J$$
 
Ltpenguin said:
Alright so i will guess that "the force" means the force from the cable so...
F=mg
F=5000 x 9.8m/s
F=49'000 N

Is that correct? or am i missing something with the 2m/s the elevator is moving. Thanks for the help.

I figured this out i believe:
F=mg+ma
F=(5000*9.8)+(5000*2)
F=59'000 N
i think that is correct. The Force placed by the tension cable to keep the elevator at rest is 49'000 Newtons to move the elevator at 2m/s it takes 59'000 N

*updated first post check it out*
 
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Still no.
 
paisiello2 said:
Still no.

What do you think it could be? i point in the right direction well help a ton thank you :)
 
If you bother to stick in the units, you'll see that they are inconsistent the way that you formulated the problem.

I think you had it on your 2nd try.
 
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  • #10
paisiello2 said:
If you bother to stick in the units, you'll see that they are inconsistent the way that you formulated the problem.

I think you had it on your 2nd try.

do you mean by 2m/s vs 9.8m/s2
 
  • #11
paisiello2 said:
I think you had it on your 2nd try.

wait so it is right?
 
  • #12
Is what right?

You gave 3 different answers. One of them is right, though.
 
  • #13
Ltpenguin said:
F=mg+ma
F=(5000*9.8)+(5000*2)
What kind of physical entity is represented by a in the first equation?
What kind of physical entity is represented by the 2 in the second equation?
 
  • #14
haruspex said:
What kind of physical entity is represented by a in the first equation?
What kind of physical entity is represented by the 2 in the second equation?

a is the acceleration of the elevator, 2m/s
 
  • #15
Ltpenguin said:
a is the acceleration of the elevator, 2m/s

2m/s is a speed, not an acceleration.
 
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  • #16
Alright so this is it
F=mg
F=5000 x 9.8m/s
F=49'000 N
The cable exerts 49k N force on the elevator, making the net force zero so it can move at a constant velocity.
 
  • #17
Yes.
 
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  • #18
w = 1 470 000 J ?
Is this correct?
 
  • #19
slland said:
w = 1 470 000 J ?
Is this correct?
Yes.
 
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