A 1.0, kg block collides with a horizontal weightless spring of force constant 2.75 Nm^{-1} as shown

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5.Work, Energy, Power and Collision

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$A$ $1.0\, kg$ block collides with a horizontal weightless spring of force constant $2.75 Nm^{-1}$ as shown in figure. The block compresses the spring $4.0\, m$ from the rest position. If the coefficient of kinetic friction between the block and horizontal surface is $0.25$, the speed of the block at the instant of collision is ................. $\mathrm{m}/ \mathrm{s}^{-1}$

A

$0.4 $

B

$4$

C

$0.8$

D

$8 $

Solution

Let the velocity of the block at the time of collision is $v$

The kinetic energy of the block is given as,

$\frac{1}{2} m v^{2}=\frac{1}{2} k x^{2}+\mu_{k} m g x$

$\frac{1}{2} \times 1 \times v^{2}=\frac{1}{2} \times 2.75 \times(4)^{2}+(0.25) \times 1 \times 9.8 \times 4$

$v^{2}=63.6$

$v=7.975 \mathrm{m} / \mathrm{s}$

Thus, the speed of block at the instant of collision is $8 \mathrm{m} / \mathrm{s}^{-1}$.

Standard 11

Physics

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