3 problems involving work and energy

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  • #1
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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.

Correct answer: 8.1 x10^4 N/m

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.

Correct answer: 5.14 m/s m/s
 
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Answers and Replies

  • #2
andrevdh
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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
jtbell
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jrd007 said:
The quation for work (W=Fdcos) confuses me.
Just what about that equation confuses you, and how?

For part b I would say you add up all the works done by forces (T, Weight of object, and Normal force?
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
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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
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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
Pyrrhus
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On problem #1
Well if you employ the work-kinetig energy theorem

[tex] \sum_{i=1}^{n} W_{i} = \Delta K [/tex]

[tex] Td - mgd = \Delta K [/tex]

were d is the magnitude of the displacement vector.

On problem #2 and #3

Use conservation fo energy

[tex] \Delta K + \Delta U = 0 [/tex]
 
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  • #7
jtbell
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jrd007 said:
The cos angle part.
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?
jrd007 said:
And to find all the forces are these the correct equations?
(F+W+N=ma)
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
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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
Doc Al
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jrd007 said:
Work of Tension = fd(cos) = 3240 N x 22 m = 71280 J
Assuming the load is raised 22 m, this is OK.
Work of Grav. = fd(cos 180) = (285 kg * 1.57) x 22 m (cos 180) =9843.9 J
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
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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
 
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  • #11
Doc Al
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jrd007 said:
(c) Now what about the work done by the cable on the load?
You already solved that one! (Post #8)
(e) Final speed = Wnet=1/2mv^2
9834 J =(1/2)(285 kg)(v^2)
v=sqr(9834J/142.5) = 8.31 m/s
Looks OK.
 
  • #12
jtbell
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OK, it looks like we can move on to problem (2) now. :smile:

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):

[tex]PE_{initial} + KE_{initial} = PE_{final} + KE_{final}[/tex]
 
  • #13
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Yes, she has mentioned that forumla.
 
  • #14
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But what does that have to do with the problem> I am clueless. Please help. :(
 
  • #15
Doc Al
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It has everything to do with Prob #2, which is a conservation of energy problem.
 
  • #16
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So do I find the force of the car? Then the work (KE) ?
 
  • #17
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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
Doc Al
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jrd007 said:
But how do I find the PE of the spring?
Look up the formula for the PE stored in a stretched (or compressed) spring.
 
  • #19
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PE = (1/2)kx^2, where K is a constant and x is the displacement?
 
  • #20
Doc Al
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That's the one. K is the "spring constant" and x is the length that the spring is stretched (or compressed).
 
  • #21
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So do I solve for my Work of Gravity done on the car? (1200 kg)(9.8 m/s^2) x (2.2 m) x cos 0. Since that is my only force can I then use KE = -PE? And then use the question below to solve for K?

PE = 1/2kx^2
 
  • #22
Doc Al
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jrd007 said:
So do I solve for my Work of Gravity done on the car? (1200 kg)(9.8 m/s^2) x (2.2 m) x cos 0.
What's the angle between the displacement of the car and the force of gravity? (It's not zero!) So, what's the work done by gravity? (Does gravity do any work if the car moves horizontally?)
Since that is my only force can I then use KE = -PE? And then use the question below to solve for K?
PE = 1/2kx^2
Use conservation of energy: Initial Energy = Final Energy. Set up the equation and solve for K.
 
  • #23
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So, there is not work being done on the car since it is moving horizotial it does not have a work for gravity or nomal force. I get that.

Oh so I would say that intial KE = Final PE

So...

1/2mv^2 = kx^2

First change velocity into m/s soo 65km/h = 65 x 1000m/3600s = 18.1 m/s

(1/2)(1200 kg)(18.1m/s)^2 = (1/2)k(2.2 m)^2
k = 196566/4.84(.5)
k = 81225 = 8.1 x 10^4 N/m

I cannot believe I got the right answer! Thanks! :)

Okay problem 3... finding intial speed.
 
  • #24
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Doc Al, how would you suggest to start problem three?

Would I use the concept that PE = mgh. But I do not know the mass of the sled.... Or maybe I could use.

PE = KE

mgh = 1/2(m)(v)^2 m's cancel out

so gh = .5v^2

v = sqr root of gh/.5

= (9.8 m/s^2 * 1.35m)/.5

= 5.14 m/s ! I got it!
 
  • #25
Doc Al
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Excellent. The moral of the story is: Don't wait until you're sure of how to solve a problem before trying something. Try various approaches until something clicks.
 

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