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A 3.628 kg freight car moving along a horizontal rail road spur track at $7.2\; km/hour$ strikes a bumper whose coil springs experiences a maximum compression of $30 \;cm$ in stopping the car. The elastic potential energy of the springs at the instant when they are compressed $15\; cm$ is [2013]
(a) $12.1 \times 10^4\;J$ (b) $121 \times 10^4\;J$ (c) $1.21 \times 10^4\;J$ (d) $1.21 \times 10^4\;J$
zero
$mgvt$ $\cos ^2 \theta$
$mgvt$ $\sin ^2 \theta$
$mgvt$ $\sin 2 \theta$
Solution
(c)
Distance $(d)$ travelled by the elevator in time $t=v t$
The block is not sliding on the wedge.
Then friction force $(f)=m g \sin \theta$ Work done by the friction force on the block in time $t$ is given by
$W=F d \cos (90-\theta)$
$\Rightarrow W= mg \sin \theta \times d \times \cos (90-\theta)$
$\Rightarrow W= mgd \sin ^2 \theta$
$\therefore W= mgvt \sin ^2 \theta$
