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Jaccobtw

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- Homework Statement
- A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle (theta) with the horizontal. A block of inertia m is pressed against the free end of the spring until the spring is compressed a distance d from its relaxed length. Call this position A. The block is the released and moves up the ramp until coming to rest at position B. The surface is rough from position A for a distance 2d up the ramp, and over this distance the coefficient of kinetic friction for the two surfaces is (μ). Friction is negligible elsewhere. What is the distance from A to B?

- Relevant Equations
- E(spring) = (1/2)kd^2

E(gravitational) = mgh

E(Work done by friction = μF(normal) *(distance)

So we know that all the energy originates from the spring:

E(spring) = (1/2)kd^2

As the block moves up the ramp, friction does work on the block over a distance of 2d:

W = μmgcos(

So subtracting the work done by friction from the spring energy, gives us the energy left, so we'll set it equal to mgh

(1/2)kd^2 - μmgcos(

Use math to get the height (h) in terms of x and sine = mgh = xmgsin(

Now solve for x:

x = ((1/2)kd^2 - 2dμmgcos(

I'm trying to figure out what I did wrong if anything.

E(spring) = (1/2)kd^2

As the block moves up the ramp, friction does work on the block over a distance of 2d:

W = μmgcos(

**θ**)* 2dSo subtracting the work done by friction from the spring energy, gives us the energy left, so we'll set it equal to mgh

(1/2)kd^2 - μmgcos(

**θ**) * 2d = mghSpring Energy - Work = Gravitational Potential EnergyUse math to get the height (h) in terms of x and sine = mgh = xmgsin(

**θ)**Now solve for x:

x = ((1/2)kd^2 - 2dμmgcos(

**θ)) / (mgsin(****θ))**I'm trying to figure out what I did wrong if anything.

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