A small block of mass \(m\) is kept at the left end of a larger block of mass \(M\) and length \(l\) . The system is started towards right with an initial velocity \(v\) . The friction coefficient between the road and the bigger block is \(k\) and that between the blocks is \(k/2\) . Find the time elapsed before the smaller block separates from the bigger block?

I am having hard time solving this , can somebody please help ,if possible with an FBD.

Thanks!

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TopNewest@Deeparaj Bhat can you please tell how did you come to know that the smaller block would move on the bigger block just by considering M>m ? – A E · 10 months, 3 weeks ago

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@Deeparaj Bhat Please help!!!! – Shivam Mishra · 11 months, 2 weeks ago

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On the bigger block, the friction at the contact surface of the smaller block is towards right (Newton's third law) and the friction at the ground surface is towards left.

Hope this helps! – Deeparaj Bhat · 11 months, 2 weeks ago

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– Shivam Mishra · 11 months, 2 weeks ago

So, will the smaller block move the total length of bigger block and then fall or it will just slip towards the left after some time interval?Log in to reply

– Deeparaj Bhat · 11 months, 2 weeks ago

It'll move the total length if \(M>m\).Log in to reply

– Shivam Mishra · 11 months, 2 weeks ago

So,for the block with mass \(M\) what would be the for net force of friction acting?Log in to reply

For the block with mass \(m\), net horizontal force (opposite to direction of \(v\)) is \(mg\dfrac{k}{2}\). – Deeparaj Bhat · 11 months, 2 weeks ago

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– Shivam Mishra · 11 months, 2 weeks ago

But how can the force on \(M\) be towards left when the system in supposed to move towards right?Log in to reply

– Deeparaj Bhat · 11 months, 2 weeks ago

Friction is always against relative motion. So, as the system goes towards right and the ground is stationary, friction acts towards left.Log in to reply

– Shivam Mishra · 11 months, 2 weeks ago

OK thanks for help!!!! I am horrible at physicsLog in to reply

– Deeparaj Bhat · 11 months, 2 weeks ago

You're welcome. :)Log in to reply