(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

One end of a spring is attached to a block and the other is attached to a wall. The spring has no mass and the block lies atop a frictionless surface. The equilibrium point of the system is at x = 0 cm. A force of 80N must be applied to the block to hold it stationary at x = -2 cm. From this position, the block is slowly moved until the applied force has done 4J of work on the spring-block system; the block is then again stationary. What is the block's position? (There are two answers.)

2. Relevant equations

-F_{spring}= F_{applied}

F_{spring}= -kx

-W_{spring}= F_{applied}

W_{spring}= 0.5k(x_{1}^{2}- x_{2}^{2})

3. The attempt at a solution

I first solved to get the spring's spring constant, k.

-F_{spring}= F_{applied}

-F_{spring}= 80N

F_{spring}= -80N

F_{spring}= -kx

-80N = -k(-0.02m)

-80N/-(-0.02m) = k

-4000N/m = k

I don't think there can be negative spring constants, so I changed this to 4000N/m.

I then solved to get the block's final position.

-W_{spring}= F_{applied}

-W_{spring}= 4J

W_{spring}= -4J

W_{spring}= 0.5k(x_{1}^{2}- x_{2}^{2})

-4J = 0.5(4000N/m)((-0.02m)^{2}- x_{2}^{2})

When I worked that equation out, I came up with a final position of +- 0.04899m, which was incorrect.

I've been at this for hours and can't solve it. Any help would be greatly appreciated.

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# Homework Help: Problem With Work Done by a Spring Force

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