# Finding Force and work on a moving object

1. May 16, 2014

### Ltpenguin

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
W=FΔd
Δd=ta

3. 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: May 16, 2014
2. May 16, 2014

### paisiello2

No.

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3. May 16, 2014

### haruspex

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?

4. May 16, 2014

### Ltpenguin

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.

5. May 16, 2014

### TrueColor

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) $$mgh = mg * 30m = 5000kg * 9.8m/s^2 * 30m = 1.47*10^6 J$$

6. May 16, 2014

### Ltpenguin

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*

Last edited: May 16, 2014
7. May 16, 2014

### paisiello2

Still no.

8. May 16, 2014

### Ltpenguin

What do you think it could be? i point in the right direction well help a ton thank you :)

9. May 16, 2014

### paisiello2

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.

Last edited: May 16, 2014
10. May 16, 2014

### Ltpenguin

do you mean by 2m/s vs 9.8m/s2

11. May 16, 2014

### Ltpenguin

wait so it is right?

12. May 16, 2014

### paisiello2

Is what right?

You gave 3 different answers. One of them is right, though.

13. May 16, 2014

### haruspex

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. May 16, 2014

### Ltpenguin

a is the acceleration of the elevator, 2m/s

15. May 16, 2014

### haruspex

2m/s is a speed, not an acceleration.

16. May 17, 2014

### Ltpenguin

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. May 17, 2014

Yes.