a 1.5-kg box is held at rest against a spring with a force constant k = 700 n/m

Question

a 1.5-kg box is held at rest against a spring with a force constant k = 700 n/m that is compressed a distance d. when the box is released, it slides across a surface that is frictionless, except for a rough patch that has a coefficient of kinetic friction Uk = 0.40 and is 6.0 cm in length. if the speed of the box is 2.6 m/s after sliding across the rough patch, determine the initial compression d of the spring.

Please include all the steps and formulas, im not just looking for the answer!

Thanks

Explanation / Answer

This is one of those "Initial potential energy = final kinetic energy" questions. But it has a little trick, the one friction patch strips away part of the final kinetic energy by the work of friction.

PEinitial = Wfriction + KEfinal

Now, solve each piece
PEinitial = 1/2 * k * x^2
Wfriction = u * N * R
(I'm gonna call the length of the friction patch R = 0.05 m)
Wfriction = u * (m * g) * R
KEfinal = (1/2) * m * v^2

PEinitial = Wfriction + KEfinal

(1/2) * k * x^2 = u * (m * g) * R + (1/2) * m * v^2

Let's get rid of the1/2 on each side

k * x^2 = 2 * u * (m * g) * R + m * v^2

Solve for x

x^2 = (2 * u * (m * g) * R + m * v^2) / k

x = SQRT((2 * u * (m * g) * R + m * v^2) / k)

x = SQRT((2 * (0.40) * ((1.5 kg) * (9.81 m/s^2) * (0.06 m) + (1.5 kg) * (2.6 m/s)^2) / (700 N/m)

x = 0.844 m

x = 8.4 cm

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