In the arrangement shown in figure, $m_{A} = 4 kg$ and $m_{B} = 1 kg$ . The system is released from rest and block $B$ is found to have a speed of $0.3 m/s$ after it has descended through a distance of $1 m$ . Find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take $g = 10 {m/s}^{2}$ ).

μ = 0.187

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μ = 0.115

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μ = 0.272

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μ = 0.323

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Solution

The correct option is B $μ = 0.115$
From constraint relations, we can say that
$V_{A} = 2 V_{B}$
Therefore, $V_{A} = 2 (0.3) m/s = 0.6 m/s$
Also if distance moved by $B$ , $S_{B} = 1 m$
$S_{A} = 2 S_{B} = 2 m$

On applying work-energy theorem
$W_{n o n - c o n s} = Δ U + Δ K . E$
where the only non conservative force acting on the system is
friction $f = μ m_{A} g$
We get: $- μ m_{A} g S_{A} = - m_{B} g S_{B} + \frac{1}{2} m_{A} V_{A}^{2} + \frac{1}{2} m_{B} V_{B}^{2}$
$⟹ - μ (4) (10) (2) = - (1) (10) (1) + \frac{1}{2} (4) (0.6)^{2} + \frac{1}{2} (1) (0.3)^{2}$
$⟹ (- 80 μ) = (- 10) + (0.72) + (0.045)$
$⟹ 80 μ = 9.235$
$⟹ μ = 0.115$

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Similar questions

Q. In the arrangement shown in figure

m_{A} = 4.0 kg

and

m_{B} = 1.0 kg .

The system is released from rest and block

B

is found to have a speed

0.3 m/s

after it has descended through a distance of

1 m

. Find the coefficient of friction between the block

A

and the table. Neglect friction elsewhere. (Take

g = 10 m / s^{2}

)

Consider the situation shown in figure. The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 $\frac{m}{s}$ after it has descended through a distance of 1m. Find the coefficient of kinetic friction between the block and the table.