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  • A block of mass 5kg and negligible dimensions are kept at the right edge of a plank of length 50cm and mass 10kg, as shown in the figure. The coefficient of friction between the plank and block is

A block of mass 5kg and negligible dimensions are kept at the right edge of a plank of length 50cm and mass 10kg, as shown in the figure. The coefficient of friction between the plank and block is 0.5 and ground is smooth. If the plank is being pulled with a force of 85N, the time taken for the block to fall over the plank is-

(g=10m/s2 )

Option: 1

1.25 sec


Option: 2

0.75 sec


Option: 3

1 sec


Option: 4

Never


Answers (1)

best_answer

Given-

Masse of block=5kg

Mass of the plank=10 kg

Coefficient of friction between blocks, \mu=0.5,

Length of Plank l=50 cm=0.5m

Only friction force is responsible for the motion of the block of mass 5 kg.

First, we need to check wheather the blockand the plank are moving together or separately for the given force of 85 N. For that, we will first assume the block and the plank are moving together with common acceleration ac. Then, we will determine the maximum possible acceleration amax of the block (when limiting friction is acting on the block). If amax is less than ac, then our assumption is right otherwise block and plank are moving separately.

F.B.D of block and plank combined-

\\ \mathrm{F_{net}=m_{sys}a}\\ \mathrm{85=15a_{c}}\\ \mathrm{a_{c}=5.67 m/s^2}

F.B.D of the block-

\\ \mathrm{N=50}\\ \mathrm{f_{l}=\mu N=25N}\\ \mathrm{f_{l}=5 a_{max}}\\ \mathrm{a_{max}=5 \ m/s^2}

Since, amax<ac

Both block and plank are moving separately. Therefore, Friction acting between the blocks will be kinetic.

Kinetic friction-

\\ \mathrm{f_{k}=\mu N=25N}\\

Let the acceleration of the plank be a1 and, acceleration of the block be a2.

F.B.D of the block and the plank-

          

For, 5 kg block-

\\ \mathrm{f_{k}=5a_{2}}\\ \mathrm{a_{2}=5\ m/s^2}

For 10 kg plank-

\\ \mathrm{85-f_{k}=10a_{1}}\\ \mathrm{a_{1}=6\ m/s^2}

In the frame of reference of the plank-

Acceleration of block with respect to plank-

\\ \mathrm{a_{21}=a_{2}-a_{1}=-1\ m/s^2}\\ \mathrm{Displacement \ of \ block\ before\ jumping\ off\ the\ plank-}\\ \mathrm{s_{21}=-0.5 m}

Let the time taken be 't' sec.

From second equation of motion -

\\ \mathrm{s_{21}=u_{21}t+\frac {1}{2} a_{12}t^2}\\ \mathrm{-0.5=-0.5t^2}\\ \mathrm{t=1\ sec}

 

Posted by

Rakesh

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