# 3 problems involving work and energy

1. Oct 13, 2005

### jrd007

Problem with Kinetic Enery
1) A 285-kg load is lifted vertically with an a = 1.060 g (1.57m/s^2), by a single cable. Determine (a) the tension in the cable (b) The net work done on the load (c) work done by the cable on the load (d) work done by gravity on the load (e) final speed of the load, assuming it started from rest.

I did a correctly gettng a tension of 3240 = 3.2 x 10^3 N

The rest I am clueless about. The quation for work (W=Fdcos) confuses me. Can anyone help me? (And yes I did draw a free-body diagram)
For part b I would say you add up all the works done by forces (T, Weight of object, and Normal force?
Correct Answers for b-e: 9.83 x 10^3 J, 7.13 x10^3 J, -6.14 x 10^4 J, 8.31 m/s

Problem involving potential energy
2) A 1200-kg car rolling on a horizontoal surface has speed v=65km/h when it strikes a horizontial coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

No idea at all how to attempt to find what the question is looking for. All I got was a conversion of v into m/s and then drew my free body picture.

Problem with Conservation of Mechanical Energy
3) A sled is intially given a shove up a frictionless 28 degree incline. It reaches a maximum vertical height 1.35 m higher than where it started. What was the intial speed?

My thoughts were to find the work done by sled in a vertical motion? Which would by F x d x cos angle. But there is no force given in the poblem or mass of the sled.

Last edited: Oct 13, 2005
2. Oct 13, 2005

### andrevdh

Apply the work-kinetic energy theorem to the second problem - that is the work done by the spring force applied to the car is equal to the change in it's kinetic energy.

3. Oct 13, 2005

### Staff: Mentor

Just what about that equation confuses you, and how?

Yes, that's one way to do it: calculate the work done by each force, separately, and then add the various amounts of work to get the net work. Another way is to add the forces to get the net force first, then calculate the work done by the net force.

4. Oct 13, 2005

### jrd007

The cos angle part.

As for # 2, you say to use W=KE=1/2(m)(v)^2

So 1/2(1200kg)(18.1 m/s)^2 = 196,566 J So then what? SInce we are looking for the force?

5. Oct 13, 2005

### jrd007

And to find all the forces are these the correct equations? (F+W+N=ma)

Nomal Force = W-T+ma = mg-T+ma then once you get the force you use Work=Fdcos?

W = -N-T+ma ?

T = W-N+ma?

6. Oct 13, 2005

### Pyrrhus

On problem #1
Well if you employ the work-kinetig energy theorem

$$\sum_{i=1}^{n} W_{i} = \Delta K$$

$$Td - mgd = \Delta K$$

were d is the magnitude of the displacement vector.

On problem #2 and #3

Use conservation fo energy

$$\Delta K + \Delta U = 0$$

Last edited: Oct 13, 2005
7. Oct 13, 2005

### Staff: Mentor

The angle is the angle between the direction of the force that you want to find the work for, and the direction the object moves in. Does that help?
Let's focus on one problem at a time, OK? In problem #1, you've already found the tension in the cable in part (a). Call it T. You already know the gravitational force, call it W, because you had to use it to find T, right? Are there any other forces acting on the load?
For part (b) you need to find the net work on the load. Pick one of the two methods given in my other posting, and try to apply it. Show your work!

8. Oct 14, 2005

### jrd007

Work of Tension = fd(cos) = 3240 N x 22 m = 71280 J

Work of Grav. = fd(cos 180) = (285 kg * 1.57) x 22 m (cos 180) =9843.9 J

But that does not add up to the correct answer. :( What am I doing wrong?

9. Oct 14, 2005

### Staff: Mentor

Assuming the load is raised 22 m, this is OK.
Two problems here: What's the force? (It's equal to the weight of the object; it's not equal to "ma".) What's cos(180)? (What is the sign of the work done by gravity?)

10. Oct 14, 2005

### jrd007

So it would be (285 kg * 9.8 m/s^s) = 2793 N

= 2973 N * 22 m * cos (180) = - 61446 J

W(net) = 71280 + (-61446) = 9834 J = 9.83 x 10^3 J

That worked! Therefore we already calulated (d) too which was -61446

(c) Now what about the work done by the cable on the load?

(e) Final speed = Wnet=1/2mv^2
9834 J =(1/2)(285 kg)(v^2)
v=sqr(9834J/142.5) = 8.31 m/s

Last edited: Oct 14, 2005
11. Oct 14, 2005

### Staff: Mentor

You already solved that one! (Post #8)
Looks OK.

12. Oct 14, 2005

### Staff: Mentor

OK, it looks like we can move on to problem (2) now.

The title of the problem gives a useful hint. I'm guessing that your book or instructor has mentioned the formula for potential energy of a compressed spring. Use conservation of total energy (PE of spring plus KE of car):

$$PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$$

13. Oct 16, 2005

### jrd007

Yes, she has mentioned that forumla.

14. Oct 16, 2005

### jrd007

15. Oct 16, 2005

### Staff: Mentor

It has everything to do with Prob #2, which is a conservation of energy problem.

16. Oct 16, 2005

### jrd007

So do I find the force of the car? Then the work (KE) ?

17. Oct 16, 2005

### jrd007

Oh, so if I find the Network/KE of the car and then the network of the spring? But how do I find the PE of the spring? I know PE = mgh but there is no h.

18. Oct 16, 2005

### Staff: Mentor

Look up the formula for the PE stored in a stretched (or compressed) spring.

19. Oct 18, 2005

### jrd007

PE = (1/2)kx^2, where K is a constant and x is the displacement?

20. Oct 18, 2005

### Staff: Mentor

That's the one. K is the "spring constant" and x is the length that the spring is stretched (or compressed).