Spring Problem On inclined plane without numbers, only variables

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SUMMARY

The discussion focuses on the energy dynamics of a block moving up an inclined plane powered by a spring. The spring's energy is defined by the equation E(spring) = (1/2)kd^2, while the work done against friction is calculated as W = μmgcos(θ) * 2d. The remaining energy is equated to gravitational potential energy, leading to the equation (1/2)kd^2 - μmgcos(θ) * 2d = mgh. The final expression for the distance x in terms of spring constant k, mass m, gravitational acceleration g, angle θ, and friction coefficient μ is derived as x = ((1/2)kd^2 - 2μmgcos(θ)) / (mgsin(θ)).

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  • Learn about the effects of friction on motion in inclined planes
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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(θ)* 2d

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(θ) * 2d = mghSpring Energy - Work = Gravitational Potential Energy

Use 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.
 
Last edited:
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Your work looks good to me.
Jaccobtw said:
Now solve for x:

x = ((1/2)kd^2 - 2μmgcos(θ)) / (mgsin(θ))
You dropped a factor of d in the second term of the numerator. But that's probably just a typing error.
 
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