Finding work done by force of friction

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Homework Help Overview

The problem involves a 4.0-kg block being lowered down a 37° incline while a horizontal force is applied. The kinetic energies at two points are given, and the task is to determine the work done by the force of friction during the movement.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the work-energy theorem and the relationship between kinetic and potential energy to analyze the problem. There are attempts to set up equations relating the energies and forces involved, with some questioning the signs and values used in their calculations.

Discussion Status

Participants are actively engaging with the problem, sharing their reasoning and calculations. Some have proposed equations to find the work done by friction, while others are clarifying their understanding of the energy transformations involved. There is no explicit consensus on the correct approach or final answer, as discrepancies in calculations have been noted.

Contextual Notes

There is mention of specific energy values and angles, as well as the need to consider the direction of forces and energy changes. Participants are also reflecting on the implications of their findings, particularly regarding the signs of the work done by friction.

Zipzap
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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?
 
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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
 
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:
 
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?
 
-fric = - 50 J (approximately)

so "fric" is 50 J. right ;-)?
 

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