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## Homework Statement

Find the amount which the spring is compressed after the mass comes to rest. Assume a kinetic coefficient of friction of 0.11 and an initial velocity along the surface of v = 4m/s.

Given values :

mass = 2.2kg

k = 500 N/m

delta = 0.036m

theta = 23 degrees

**The attempt at a solution**

Alright.. here goes

My approach to this problem is to find the velocity at where the block touches the spring, and make it as the initial velocity.

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Using a free body diagram, I got the following values.

F = mg = 21.56N

N = mg cos theta = (21.56)cos23 = 19.846N

friction force = N.coefficient = (19.846)(0.11) = 2.183N

force down the plane = mg sin theta = (21.56)sin23 = 8.424N

Adding up the force in the x direction (parallel to the slope), F = 8.424 + (-2.183) = 6.241N

since F = ma,

6.241N = (2.2kg)a

therefore, a = 2.837m/s^2 [sorry, i'm not familiar with the forums. the unit is metre per seconds squared]

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Now, with the acceleration, the velocity at when it touches the spring can be calculated :

final velocity squared = initial velocity squared + 2(acceleration)(distance)

v^2 = u^2 + 2(a)(r)

v^2 = 4^2 + 2(2.837)(0.036)

v = squareroot(16 + 0.204)

v = 4.025m/s

therefore, the velocity at when it touches the spring is 4.025m/s

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Using the work-energy equation,

T1 + U = T2

as when the final velocity is when the block comes to rest, the final velocity is 0, therefore

(mv^2)/2 + U = 0

17.821 +U = 0

therefore U = 17.821J

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Using the value, the compression length can be found:

17.821 = 0.5(k)(final compression length squared - initial compression length squared)

17.821 = 0.5(500)(x^2 - 0)

17.821 = 250(x^2)

0.071283 = x^2

x = squareroot(0.071283)

**x = 0.267m**

that was my answer, but it's wrong. Much help would be appreciated on where i went wrong and how to solve the problem. Thanks.