Waste less time on Facebook — follow Brilliant.
×

Friction doubt!

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!

Note by Shivam Mishra
3 months, 1 week ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

@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 ? Albert Einstein · 2 months, 2 weeks ago

Log in to reply

@Deeparaj Bhat Please help!!!! Shivam Mishra · 3 months, 1 week ago

Log in to reply

@Shivam Mishra On the smaller block, the friction is towards the left (assuming the velpcity is towards the right).

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 · 3 months, 1 week ago

Log in to reply

@Deeparaj Bhat 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? Shivam Mishra · 3 months, 1 week ago

Log in to reply

@Shivam Mishra It'll move the total length if \(M>m\). Deeparaj Bhat · 3 months, 1 week ago

Log in to reply

@Deeparaj Bhat So,for the block with mass \(M\) what would be the for net force of friction acting? Shivam Mishra · 3 months, 1 week ago

Log in to reply

@Shivam Mishra For the block with mass \(M\), Net force in horizontal direction is given by \(Mgk-mg\dfrac{k}{2}\) towards the left (opposite to direction of \(v\) ).

For the block with mass \(m\), net horizontal force (opposite to direction of \(v\)) is \(mg\dfrac{k}{2}\). Deeparaj Bhat · 3 months, 1 week ago

Log in to reply

@Deeparaj Bhat But how can the force on \(M\) be towards left when the system in supposed to move towards right? Shivam Mishra · 3 months, 1 week ago

Log in to reply

@Shivam Mishra Friction is always against relative motion. So, as the system goes towards right and the ground is stationary, friction acts towards left. Deeparaj Bhat · 3 months, 1 week ago

Log in to reply

@Deeparaj Bhat OK thanks for help!!!! I am horrible at physics Shivam Mishra · 3 months, 1 week ago

Log in to reply

@Shivam Mishra You're welcome. :) Deeparaj Bhat · 3 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...