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

A block of mass m rests on a plane inclined at an angle θ with the horizontal. A spring with force constant k is attached to the block. The coefficient of static friction between the block and plane is μ_{s}. The spring is pulled upward along the plane very slowly.

What is the extension of the spring the instant the block begins to move. (Use any variable or symbol stated above along with the following as necessary: g.)

2. Relevant equations

F = ma

F = -kx

3. The attempt at a solution

I figured I can sum up forces in X and Y, with X being in the direction of the plane, and solve for x, but it's coming back as incorrect. I tried reversing the signs in case I missed a negative as well. The only other option I can see is summing the forces in the X direction equal to zero which would remove the ma term, but the system could be in equilibrium for any value of X and the block still not move. My thoughts are that setting Fx=ma finds x the instant it moves. I only have one submission left so I wanted to check before I go any farther.

ƩF_{y}=F_{N}-mgcos(θ)=0

F_{N}=mgcos(θ)

ƩF_{x}=-F_{N}μ_{s}-mgsin(θ)+kx=ma

Substitute F_{N}

-mgcos(θ)μ_{s}-mgsin(θ)+kx=ma

Solving for x

(-mgcos(θ)μ_{s}-mgsin(θ)-ma)/k=x

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# Homework Help: Potential Energy, Springs, and Friction

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