Finding work done by force of friction

In summary, the block does work against the force of friction between A and B, and the amount of work done is -68 J.
  • #1
Zipzap
34
0

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?
 
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  • #2
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
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
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
-fric = - 50 J (approximately)

so "fric" is 50 J. right ;-)?
 

1. What is the formula for calculating work done by force of friction?

The formula for calculating work done by force of friction is W = μ * N * d, where W is work, μ is the coefficient of friction, N is the normal force, and d is the displacement.

2. How does the coefficient of friction affect the work done by force of friction?

The coefficient of friction determines the amount of force required to overcome friction and move an object. The higher the coefficient of friction, the more work is done by the force of friction.

3. Is the work done by force of friction always negative?

Yes, the work done by force of friction is always negative because the force of friction acts in the opposite direction of the object's movement, therefore reducing the object's kinetic energy.

4. How does the normal force affect the work done by force of friction?

The normal force is perpendicular to the surface and determines the amount of frictional force between the two surfaces. The greater the normal force, the greater the work done by force of friction.

5. Can the work done by force of friction be greater than the work done by the applied force?

No, the work done by force of friction cannot be greater than the work done by the applied force. The work done by friction is always less than the work done by the applied force due to the loss of energy through friction.

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