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Question: A block of mass m is moving with a speed v on a horizontal rough surface and collides with a horizon...

A block of mass m is moving with a speed v on a horizontal rough surface and collides with a horizontally mounted spring of spring constant k as shown in the figure. The coefficient of friction between the block and the floor is μ\mu . The maximum compression of the spring is :

A

−μmgk+1k(μmg)2+mkv2- \frac { \mu \mathrm { mg } } { \mathrm { k } } + \frac { 1 } { \mathrm { k } } \sqrt { ( \mu \mathrm { mg } ) ^ { 2 } + \mathrm { mkv } ^ { 2 } }

B

μmgk+1k(μmg)2−mkv2\frac { \mu \mathrm { mg } } { \mathrm { k } } + \frac { 1 } { \mathrm { k } } \sqrt { ( \mu \mathrm { mg } ) ^ { 2 } - \mathrm { mkv } ^ { 2 } }

C

−μmgk−1k(μmg)2−mkv2- \frac { \mu \mathrm { mg } } { \mathrm { k } } - \frac { 1 } { \mathrm { k } } \sqrt { ( \mu \mathrm { mg } ) ^ { 2 } - \mathrm { mkv } ^ { 2 } }

D

μmgk+1k(μmg)2+mkv2\frac { \mu \mathrm { mg } } { \mathrm { k } } + \frac { 1 } { \mathrm { k } } \sqrt { ( \mu \mathrm { mg } ) ^ { 2 } + \mathrm { mkv } ^ { 2 } }

Answer

−μmgk+1k(μmg)2+mkv2- \frac { \mu \mathrm { mg } } { \mathrm { k } } + \frac { 1 } { \mathrm { k } } \sqrt { ( \mu \mathrm { mg } ) ^ { 2 } + \mathrm { mkv } ^ { 2 } }

Explanation

Solution

In presence of friction, both the spring force and the frictional act so as to oppose the

compression of the spring.

Work done by the net force

W=−12kxm2−μmgxm\mathrm { W } = - \frac { 1 } { 2 } \mathrm { kx } _ { \mathrm { m } } ^ { 2 } - \mu \mathrm { mgx } _ { \mathrm { m } }

Where XmX _ { m } is the maximum compression of the spring. Change in kinetic energy.

ΔK=kf−ki=0−12mv2\Delta \mathrm { K } = \mathrm { k } _ { \mathrm { f } } - \mathrm { k } _ { \mathrm { i } } = 0 - \frac { 1 } { 2 } \mathrm { mv } ^ { 2 }

According to work-energy theorem

W=Δk\mathrm { W } = \Delta \mathrm { k }

−12kxm2−μmgxm=−12mv2- \frac { 1 } { 2 } \mathrm { kx } _ { \mathrm { m } } ^ { 2 } - \mu \mathrm { mgx } _ { \mathrm { m } } = - \frac { 1 } { 2 } \mathrm { mv } ^ { 2 }

12kxm2+μmgxm=−12mv2\frac { 1 } { 2 } \mathrm { kx } _ { \mathrm { m } } ^ { 2 } + \mu \mathrm { mgx } _ { \mathrm { m } } = - \frac { 1 } { 2 } \mathrm { mv } ^ { 2 }

kxm2+2μmgxm−mv2=0\mathrm { kx } _ { \mathrm { m } } ^ { 2 } + 2 \mu \mathrm { mgx } _ { \mathrm { m } } - \mathrm { mv } ^ { 2 } = 0

xm2+2μmgxmk−mv2k=0\mathrm { x } _ { \mathrm { m } } ^ { 2 } + \frac { 2 \mu \mathrm { mgx } _ { \mathrm { m } } } { \mathrm { k } } - \frac { \mathrm { mv } ^ { 2 } } { \mathrm { k } } = 0

It is a quadratic equation in xm\mathbf { x } _ { \mathrm { m } } .

Solving this quadratic equation for and taking only positive root since is positive, we get