Question

The spring is compressed by a distance A on a rough surface and the system is released. The block again comes to an instantaneous rest when the spring is elongated by a distance B during the process find work done and coefficient of friction.... kindly solve it ASAP

Answer 1

Answer:

Work done $=\frac{1}{2}k(a^2-b^2)$

Coefficient of friction $=\frac{k(a-b)}{2mg}$

Explanation:

If the spring constant is k and mass of the block is m

when the spring is compressed by a the restoring force, the work done

$W=\frac{1}{2}ka^2$

When the spring is stretched by b, the work done

$W'=-\frac{1}{2}kb^2$

Total work done

$=\frac{1}{2}k(a^2-b^2)$

Total work done by the friction force

$F=\mu\times mg\times (a+b)$

This work done will be equal to the work done by on the spring

Thus

$\mu\times mg\times (a+b)=\frac{1}{2}k(a^2-b^2)$

$\implies \mu=\frac{1}{2}\times\frac{k(a+b)(a-b)}{mg(a+b)}$

$\implies \mu=\frac{k(a-b)}{2mg}$

Hope this answer is helpful.