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A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring distance of 0.20m (fig). When released, the block moves on a horizontal table top for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction, $μ_{K}$ between the block and the table?

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0.50

0.25

0.40

None of these

answer is A.

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Detailed Solution

The initial and final kinetic energies are both zero, so the work done by the spring is the negative of the work done by friction, or $\frac{1}{2} {kx}^{2} = μ_{k} mgl,$ where l is the distance the block moves. Solving for $μ_{k}$ ,

$μ_{k} = \frac{(\frac{1}{2}) {kx}^{2}}{mgl} = \frac{(\frac{1}{2}) (100 N / m) {(0.20 m)}^{2}}{(0.50 kg) (10 m / s^{2}) (1.00 m)} = 0.40$

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