# Work Energy Theorem Problem

1. Feb 8, 2013

### TheRedDevil18

1. The problem statement, all variables and given/known data
I do not know how to find the height. I am thinking of using mgh=delta k but I am not getting the correct answer. I know the answer is 6.28m because my teacher gave it to me.
Below is the picture and question:

2. Relevant equations
mgh = delta k

3. The attempt at a solution
ep=ek
mgh=785000
98000h=785000
h = 8.01m

2. Feb 8, 2013

### Staff: Mentor

Since the truck moves at constant speed, ΔKE = 0. So that won't work.

Instead, consider the work done by the engine and where that work goes.

3. Feb 8, 2013

### sankalpmittal

Take truck as the system. What force acts during its ascension ? How many are non conservative ? What is work done by non conservative force ?

Negative work done by non conservative force= Change in truck's total mechanical energy.

4. Feb 8, 2013

### TheRedDevil18

The non conservative force will be the friction right? so if I set up the equation it will be like this
-85000=k1+u1

5. Feb 8, 2013

### sankalpmittal

I should have told you to take truck+earth as the system to account for external forces.

Yes. Now total change in mechanical energy is change in kinetic energy plus change in potential energy. As change in kinetic energy is zero, so what is the change in potential energy? It will be better to account for minus sign to gravity.

6. Feb 8, 2013

### TheRedDevil18

mgh, Cant work it out, need the height.

7. Feb 8, 2013

### sankalpmittal

Wfriction = M.Ef-M.Ei

M.E. is mechanical energy..

Wfriction = ΔK.E + ΔP.E.

As ΔK.E=0 , so

ΔPE=mgh

Now you know, Wfriction= -85000 J.. Can you work it out from here ?

8. Feb 8, 2013

### Staff: Mentor

The height is what you are supposed to find! Just call it h and you will solve for it.

9. Feb 9, 2013

### TheRedDevil18

Okay I think I got the answer:
Fapp-Ffric=mgh
7*10^5-8.5*10^4=10000*9.8*h
h = 6.28m

Correct?

10. Feb 9, 2013

### Staff: Mentor

Yes, correct. (Where Fapp is the work done by the applied force, the engine in this case. Better to call it Wapp, right? And the work done against friction would be Wfric.)

11. Feb 9, 2013

### TheRedDevil18

Yeah, sorry its work not force. Anyway,for 3.2 what do they mean by the instantaneous power delivered by the truck. I know P=w/t and I got the time to be 4.14s using t=d/s but what should I use for work, must I subtract Wapp-Wfrict and put it into the equation?

12. Feb 9, 2013

### Staff: Mentor

Good.
To find the power delivered by the engine, just use the work done by the engine.

13. Feb 9, 2013

### TheRedDevil18

Okay so,
p=w/t
= 7*10^5/4.14
= 169082.13 W

Correct?

14. Feb 9, 2013

### Staff: Mentor

Good. (I suggest rounding off your final answer and using exponential notation.)

15. Feb 9, 2013

### TheRedDevil18

Will do, and thanks for your help