A 4 kg block is kept on to a long 12 kg block, which is accelerating along a smooth horizontal table at a = 5.2 m / s 2 under application of an external constant force. Coefficient of friction between

Question

A 4 kg block is kept on to a long 12 kg block, which is accelerating along a smooth horizontal table at a = 5.2 m / s 2 under application of an external constant force. Coefficient of friction between the two blocks is half the minimum value which will prevent the relative slipping between 4 kg and 12 kg blocks. If the amount of heat generated due to sliding between the two blocks during the time in which 12 kg block moves 10 m starting from rest is α J. Find the value of 13 α ​ .

TestHub · Accepted Answer

Minimum friction coefficient to prevent slipping: the 4 kg block needs $f = 4 \times 5.2 = 20.8 N$ , so $μ_{min} = 20.8/40 = 0.52$ ; actual $μ = 0.26$ .

Friction force 10.4 N gives the 4 kg block $a_{1} = 2.6 m/ s^{2}$ ; relative acceleration = 2.6 m/s². While the 12 kg block covers 10 m, $t^{2} = 20/5.2$ , and relative displacement $= \frac{1}{2} \times 2.6 \times t^{2} = 5 m$ .

$α = Q = f \times 5 = 10.4 \times 5 = 52 J \Rightarrow α /13 = 4$

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