Work Done By Friction problem

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  • #1
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"What is the work doen by friction on a 15 kg object pulled horizontally in a straigth line for 15 meters, if the coefficient of friction between the object and the surface is given by uk = .04?"

a) -59 J
b) -91 J
c) -145 J
d) -590 J
e) -890 J

Im not sure how I would solve this problem. From what I've read, and was given, "the work done on an object is equal to the product of the parallel component of force and the magnitude of displacement it causes."

well, the displacement was 15m, but the component of the force is what im having trouble with. The normal force is 147, multiplied by .04 would give me 58.8. This would lead me to choice A, however, Im sure im missing some important steps.
 

Answers and Replies

  • #2
rsk
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Well, I haven't managed to get it to equal any of those values exactly, but remember that the 58.8 you've calculated is the value of the frictional force, not the work done by it.
 
  • #3
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one of the things I was trying is that by calculating the product of the component of the force (58.8) and the 15m displacement, which gives me 882. Its close to 890, but I still doubt its the answer.
 
  • #4
rsk
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I also think you might be a factor of 10 out.

Unless I am also missing something:

Weight = mg so Normal also = mg = 15 x 9.81 = 147 as you say. SO Friction = 0.04 of this = 5.89

So Work done by friction is this x the distance which I get to be -88.29.

Now, using g = 10 instead of 9.81 gives 90 exactly, so I don't understand why it's 91. So, I may be utterly misunderstanding this - hopefully someone else can confirm it for you.
 
  • #5
nrqed
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I also think you might be a factor of 10 out.

Unless I am also missing something:

Weight = mg so Normal also = mg = 15 x 9.81 = 147 as you say. SO Friction = 0.04 of this = 5.89

So Work done by friction is this x the distance which I get to be -88.29.

Now, using g = 10 instead of 9.81 gives 90 exactly, so I don't understand why it's 91. So, I may be utterly misunderstanding this - hopefully someone else can confirm it for you.

I agree with all those numbers. I also thought for a second that they had used g =10 m/s^2 instead of the usual 9.80 m/s^2 but it does give -90 J in that case

(to the OP: the work done by the friction force is [itex] \mu_k n d cos(180) = -\mu_k n d [/itex].)

I don't know how they could have got -91 J but that's the closest.
 
  • #6
PhanthomJay
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Please check the problem statement again. I have a sneaky feeling that the friction coefficient is 0.4, not the 0.04 that you have written. Just a typo, that's all. Since the friction force on the object is is in the opposite direction of its displacement, the work will be negative.
[tex] W_f = =-u_kNd = -0.4(15)(9.8)(15) = -882J[/tex]. That'd makes (e) the correct answer. Why do you doubt your work?
 
  • #7
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hey, you're right Phanthom, it is 0.4. Sorry for the mistake guys!
As for your question, I dont know why I doubt my work, just trying to make sure Im going in the right direction. Thanks again guys
 

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