Question

An ideal massless spring $S$ can be compressed $1.0m$ in equilibrium by a force of $100N$ . This same spring is placed at the bottom of a friction less inclined plane which makes an angle $\theta =30^{\circ }$ with the horizontal. A $10kg$ mass $m$ is released from the rest at the top of the inclined plane and is brought to rest momentarily after compressing the spring by $2.0m$ . What will be the distance through which the mass moved before coming to rest ?

• A. $4m$
• B. $6m$
• C. $8m$
• D. $5m$

Answer 1

Answer: (C)

Given,
The angle of inclination is, $\theta =30^{\circ }$
The spring got compressed by, $l=2m$
The mass of the body, $m=10kg$
When a force, $F=100N$ is applied, the spring gets compressed by $x=1m$ .
Therefore, the spring constant, $k=\frac{F}{x}=\frac{100}{1}=100N/m$
The net force acting on the body along the incline of the body is given by, $F_i=mgsin\theta =10\times 10\times sin30=50N$
Say, the body has travelled a vertical distance $d$ .
So, the work done by the gravity is, $W_g=F_i\times d=50d$
The work done by the spring is given by, $W_s=\frac{1}{2}k{l}^2=\frac{1}{2}100\times 2^2=200J$
As the spring undergoes maximum compression, its speed becomes zero. So, the change in its kinetic energy is zero. Therefore, from work energy theorem,
$W_g=W_s\Rightarrow 50d=200\Rightarrow d=4m$
The distance travelled by body, $s=\frac{d}{sin30^{\circ }}=8m$