Q. The spring is compressed by a distance $a$ and released. The block again comes to rest when the spring is elongated by a distance $b$. During this :

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A

Work done by the spring on the block $=\frac{1}{2} k(a+b)^{2}$

B

Work done by the spring on the block $=\frac{1}{2} k\left(a^{2}-b^{2}\right)$

C

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

D

Coefficient of friction $=\frac{k(a+b)}{2 m g}$

Solution:

Work done by spring on block,

$w_{s}=\frac{1}{2} k a^{2}-\frac{1}{2} k b^{2}=\frac{1}{2} k\left(a^{2}-b^{2}\right)$
By work energy theorem,
$K_{i}+w =K_{f} $
$0+\frac{1}{2} k a^{2}-\mu m g(a+b)-\frac{1}{2} k b^{2}=0$
$\mu m g(a+b) =\frac{1}{2} k\left(a^{2}-b^{2}\right)$
$\mu =\frac{k(a-b)}{2 m g}$

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Q. The spring is compressed by a distance $a$ and released. The block again comes to rest when the spring is elongated by a distance $b$. During this :

Work done by the spring on the block $=\frac{1}{2} k(a+b)^{2}$

Work done by the spring on the block $=\frac{1}{2} k\left(a^{2}-b^{2}\right)$

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

Coefficient of friction $=\frac{k(a+b)}{2 m g}$

Work done by spring on block, $w_{s}=\frac{1}{2} k a^{2}-\frac{1}{2} k b^{2}=\frac{1}{2} k\left(a^{2}-b^{2}\right)$ By work energy theorem, $K_{i}+w =K_{f} $ $0+\frac{1}{2} k a^{2}-\mu m g(a+b)-\frac{1}{2} k b^{2}=0$ $\mu m g(a+b) =\frac{1}{2} k\left(a^{2}-b^{2}\right)$ $\mu =\frac{k(a-b)}{2 m g}$

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4. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c}=k^{2} \, rt^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

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Latest Updates

Work done by spring on block,

$w_{s}=\frac{1}{2} k a^{2}-\frac{1}{2} k b^{2}=\frac{1}{2} k\left(a^{2}-b^{2}\right)$
By work energy theorem,
$K_{i}+w =K_{f} $
$0+\frac{1}{2} k a^{2}-\mu m g(a+b)-\frac{1}{2} k b^{2}=0$
$\mu m g(a+b) =\frac{1}{2} k\left(a^{2}-b^{2}\right)$
$\mu =\frac{k(a-b)}{2 m g}$