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Q. A $1\, kg$ block collides with a horizontal weightless spring of force constant $2.75 \,N / m$ as shown in figure. The block compresses the spring $4 \,m$ from the free position. If the coefficient of kinetic friction between the block and the horizontal surface is $0.25$, find the speed(in $m / s )$ of the block at the instant of collision in $m / s$. $\left( g =10 \,m / s ^{2}\right)$Physics Question Image

Work, Energy and Power

Solution:

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$\therefore W$ by friction $+ W$ by Spring $=0-\frac{1}{2} mV ^{2}$
$-\mu mgx -\frac{ K }{2} x ^{2}=-\frac{1}{2} mV ^{2} $
$0.25 \times 1 \times \frac{10 \times 4}{100}+\frac{2.75}{2} \times \frac{16}{1000}=\frac{1}{2} \times 1 \times V ^{2} $
$ 10 +2.75 \times 8=\frac{ V ^{2}}{2} $
$ 32 =\frac{ V ^{2}}{2} $
$ V =8 \,m / s $