Energy loss: trouble with signs?

[SbCl3]

Homework Statement

picture: http://is.gd/tm7a

An apparatus to determine coefficients of friction is shown above. The box is slowly rotated counterclockwise. When the box makes an angle $$\theta$$ 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 $$\theta$$, 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:

$$\Delta$$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($$\theta$$)
potential energy due to the spring: .5*k*x²

The Attempt at a Solution

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


All I don't understand are the signs. Does not $$\Delta$$E = E_final - E_initial? (in other words, the change is always final minus initial)

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

Help is greatly appreciated.

 

[Astronuc]

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.

 

[nlightner]

You are correct in your understanding that \DeltaE = Efinal - Einitial. The given solution may appear to have incorrect signs, but this is due to the reference frame being used. In this problem, the initial energy is taken as zero, so the final energy will always be less than zero (since the block loses energy as it slides down the incline). Therefore, the expression \DeltaE = mg(d+x)sin(theta) - 1/2*k*x2 can be rewritten as \DeltaE = -mg(d+x)sin(theta) + 1/2*k*x2, which follows the same principle of final energy being less than initial energy. This is just a matter of convention and does not affect the final result. So, your attempt at a solution is also correct.