just some Friction!

Classical Mechanics Level 3

Blocks A and B each of mass 1 kg are moving with 4m s − 1 s^{-1} and 2m s − 1 s^{-1} respectively,as shown. The coefficient of friction for all surface is 0.10. Find the distance (in m) by which centre of mass will travel before coming to rest. Assume large distance between the two blocks.

(Assume that g = 10 m/s^2.)

5 3 8 4

Prashant Kr by Prashant Kr , 22, India

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2 solutions

Arjen Vreugdenhil
Sep 21, 2015

a = F m = μ m g m = μ g = 0.10 ⋅ 10 = 1.0 m/s 2 . a = \frac{F}{m} = \frac{\mu m g}{m} = \mu g = 0.10\cdot 10 = 1.0\:\text{m/s}^2. Δ x A = 0 2 − v A 2 2 a A = − 4 2 2 ⋅ ( − 1 ) = + 8 m . \Delta x_A = \frac{0^2-v_A^2}{2a_A} = \frac{-4^2}{2\cdot (-1)} = +8\:\text{m}. Δ x B = 0 2 − v B 2 2 a B = − ( − 2 ) 2 2 ⋅ 1 = − 2 m . \Delta x_B = \frac{0^2-v_B^2}{2a_B} = \frac{-(-2)^2}{2\cdot 1} = -2\:\text{m}. Δ x C M = m A Δ x A + m B Δ x B m A + m B = ( + 8 ) + ( − 2 ) 2 = 3 m . \Delta x_{CM} = \frac{m_A\:\Delta x_A + m_B\:\Delta x_B}{m_A + m_B} = \frac{(+8)+(-2)}{2} = 3\:\text{m}.

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Prashant Kr
Apr 8, 2015

Retardation of each block =1 m s − 2 s^{-2} . Distance travelled by A before it stops : x 1 x_{1} =8 m.

Distance travelled by B before it stops: x 2 x_{2} =2m.

Change in X c m X_{cm} = 3 m

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isnt net force on the system zero,bro as the friction forces are equal and opposite..

Chirag Shyamsundar - 6 years ago

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The net force on the system is zero during the first two seconds. After that, block B is standing still, and the net force is to the left due to friction on block A only.

An alternative analysis is therefore: Have the CM travel at constant velocity of 1 m/s during the first two seconds, then slow down to a full stop in the next two seconds. Distance traveled: 2 m + 1 m.

Arjen Vreugdenhil - 5 years, 8 months ago

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