Question:
Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass $1.0 \mathrm{~kg}$ is found to have a speed $0.3 \mathrm{~m} / \mathrm{s}$ after it has descended through a distance of $1 \mathrm{~m}$. Find the coefficient of kinetic friction between the block and the table.
Solution:
Work done = change in K.E.
$-\mu R S+m g=\left[^{\frac{1}{2}} m_{1} v_{1}^{22} m_{2} v_{2}^{2}\right]-0$$-\mu \times 40 \times 2+1 \times 10=\frac{1}{2}\left(4 \times 0.3^{2}+1 \times 0.6^{2}\right)$
$-\mu \times 40 \times 2+1 \times 10=\frac{1}{2}\left(4 \times 0.3^{2}+1 \times 0.6^{2}\right)$
$\mu=0.12$
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