Question

An ideal massless spring can be compressed by $1m$ by a force of $100N$. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle $\theta ={30}^o$ with the horizontal. A $10kg$ mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring $2$ meters
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?

Answer 1

(a) Let total distance moved by the block is $S=(l+2)m$ where $l$ is the distance moved by the block before touching the spring
Now, work done by gravity on the block is
$W_g=mgS\sin \theta =10\times 10\times S\sin 30J\Rightarrow W_g=50SJ$
Work done by spring on the block is, $W_S=-\frac{1}{2}k{x}^2\Rightarrow W_S=-\frac{1}{2}\times 100\times 4J=W_S=-200J....\left(2\right)$
Total work done $W=W_g+W_S\Rightarrow W=(50S-200)J$
Since change in KE of the block is zero as $W=\Delta K.E\Rightarrow 50S-200=0\Rightarrow S=4m$

(b) As $S=l+2\Rightarrow l=S-2=2m$

Work done by gravity over this path length is
$W_g=mg\times 2\sin \theta =10\times 10\times 2\times \frac{1}{2}=100J[\because W_g=\Delta K.E]$

$\Rightarrow 100=\frac{1}{2}m{v}^2-0=v^2=\frac{100\times 2}{10}=20\Rightarrow v=2\sqrt{5}m/s$