Q. 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 ?
NTA AbhyasNTA Abhyas 2020
Solution:
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, $W_{s}=\frac{1}{2}kl^{2}=\frac{1}{2}100\times 2^{2}=200J$
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, $s=\frac{d}{sin 30 ^\circ }=8m$
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, $W_{s}=\frac{1}{2}kl^{2}=\frac{1}{2}100\times 2^{2}=200J$
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, $s=\frac{d}{sin 30 ^\circ }=8m$