Solve the following :

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)$

$-\mu \times 40 \times 2+1 \times 10=\frac{1}{2}\left(4 \times 0.3^{2}+1 \times 0.6^{2}\right)$

$\mu=0.12$

Leave a comment

None
Free Study Material