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Energy loss: trouble with signs?

  • Thread starter SbCl3
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Homework Statement



picture: http://is.gd/tm7a [Broken]

An apparatus to determine coefficients of friction is shown above. The box is slowly rotated counterclockwise. When the box makes an angle [tex] \theta [/tex] with the horizontal, the block of mass m just starts to slide, and at this instant the box is stopped from rotating. Thus at angle [tex]\theta[/tex], the block slides a distance d, hits the spring of force constant k, and compresses the spring a distance x before coming to rest. In terms of the given quantities, derive an expression for each of the following:

[tex]\Delta[/tex]E, the loss in total mechanical energy of the block‑spring system from the start of the block down the incline to the moment at which it comes to rest on the compressed spring.

Homework Equations



potential energy due to gravity: mgh or mg(x+d)sin([tex]\theta[/tex])
potential energy due to the spring: .5*k*x2

The Attempt at a Solution



The answer given is: [tex]\Delta[/tex]E = mg(d+x)sin(theta) - 1/2*k*x2


All I don't understand are the signs. Does not [tex]\Delta[/tex]E = Efinal - Einitial? (in other words, the change is always final minus initial)

If that is the case, [tex]\Delta[/tex]E = 1/2*k*x2 - mg(x+d)sin([tex]\theta[/tex]).

Help is greatly appreciated.
 
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Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
18,704
1,718
The problem asks for "the loss in total mechanical energy of the block‑spring system".

The gravitational potential is decreasing (hence + loss, or - gain), but the spring is being compressed, to it's mechanical energy is increasing (- loss, or + gain). The difference between the two, the loss, is dissipated by friction.
 

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