Find the work done

  • Thread starter dolpho
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  • #1
dolpho
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Homework Statement



In the system shown in the figure, suppose the block has a mass of 2.7kg , the spring has a force constant of 540 n/m , and the coefficient of kinetic friction between the block and the floor is 0.16.

Find the work done on the block by the spring and by friction as the block is moved from point A to point B along path 2.

http://imgur.com/Rr7lE

Homework Equations



Work = 1/2 mv^2

Work by spring = 1/2 kx^2

Friction work = -Ukmgx

The Attempt at a Solution



For the first part I did; 1/2 (540)(.02m^2) = .108

For the work of friction I used Friction work = -ukmgx = -(.16)(2.7)(9.81)(.02^2) = -.085

I've also tried, -(.16)(2.7)(9.81)(.06^) = -.015

I'm not sure what I'm doing wrong but MasteringPhysics is saying check my sign on the first term, I added a negative to the first and then it said it was all wrong and didn't give me any hints. Any idea what I'm doing wrong?
 

Answers and Replies

  • #2
haruspex
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What is the force in the spring initially (point A)? What is it when the block is at point B? What is the average force over that distance?
Btw, you wrote -(.16)(2.7)(9.81)(.02^2), but you obviously didn't mean the ^2.
 

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