Finding work done by force of friction

  • Thread starter Zipzap
  • Start date
  • #1
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Homework Statement


A 4.0-kg block is lowered down a 37° incline a distance of 5.0 m from point A to point B. A horizontal force (F = 10 N) is applied to the block between A and B as shown in the figure. The kinetic energy of the block at A is 10 J and at B it is 20 J. How much work is done on the block by the force of friction between A and B?

(I can't find a picture of this, but the block is being lowered down an incline going right, and the horizontal force is directly left, parallel to the ground)

Homework Equations



Ek = 0.5mv^2
Ep = mgh
(delta)E = Ef - Ei

The Attempt at a Solution



After some research, the answer is supposed to be -68 J

However, I'm stuck as to how I have to approach this problem. My initial idea was to determine the potential and kinetic energies of both A and B, and somehow use basic math and plug n' chug to get my answer. Is this the right approach?
 

Answers and Replies

  • #2
60
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You know the initial KE of the block at "A". You also know the kinetic energy of the block at "B". The reduction in the potential energy (m*g* 5 * sin(37)) must cause in increase in KE of the block. The frictional force will dissipate the KE of the block and the component of the applied force along the incline (10 * cos(37)) to the left will dissipate the KE too.

So

KE(A) = 10

KE(B) = KE(A) + (m*g* 5 * sin(37)) - Frictional energy - (energy lost due to applied force along the incline tending to push it up along the incline ie - (10 * cos(37)) * 5

Plug in the numbers and get the Frictional energy
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi Zipzap! :wink:

(have a delta: ∆ and try using the X2 and X2 icons just above the Reply box :wink:)

Use the work energy theorem …

total work done = change in mechanical energy …

what do you get? :smile:
 
  • #4
34
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Venkatg: Let me see if I have this down right...

KE(B) = KE(A) + PE(A) - F(fric) - W(app)

-F(fric) = KE(B) - KE(A) - PE(A) + W(app)

-F(fric) = 20 - 10 - 4*9.8*5sin37 + 10*5*cos37

Unfortunately, I get the right number, but I end up getting it positive and not negative. Where did I go wrong?
 
  • #5
60
0
-fric = - 50 J (approximately)

so "fric" is 50 J. right ;-)?
 

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