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Question

Coefficient of friction between two blocks shown in figure is μ=0.6 The blocks are given velocities in the direction shown in figure. Find the displacements of 1kg and 2kg blocks upto that instant. (Take g= 10m/s2)
251105_d1895d1a130c4f3badfcb0fefed4a66e.png

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Solution

With respect to 1kg block, 2kg block is moving in backward direction at velocity of, vr=3+18=21m/s, in the forward direction.

frictional force on 2kg block, in the backward direction is, f=μN=0.6×2kg×10m/s2=12N
So, acceleration of 2kg block in ground's frame is, a2=f/2kg=6ms2, in the backward direction.

Similar frictional force applies on 1kg block in the forward direction.
So, deceleration of 1kg block is, a1=f/1kg=12m/s2, in the forward direction.

With respect to 1kg block, there is pseudo force of 2kg×a1=24N, on 2kg block, in the backward direction.
So, acceleration of 2kg block with respect to 1kg block is, ar=(24N+12N)/2kg=18m/s2, in the backward direction.
So time taken to stop relative motion is, t=vr/ar=21ms1/18ms2=1.167s

Now, displacement of 1kg block and 2kg block are given as,
s1=18t+12a1t2=18t+1212t2
s2=3t12a2t2=3t126t2

At t=1.167s,
s1=776m
s2=712m

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