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 $2.0m$. The distance through which the mass moved before coming to rest is:

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

Answer 1

Answer: (C)

$4m$

Spring is compressed 1 m with the action of $100N$ force

from this we can calculate the spring constant k.

$F=kx100=k\times 1k=100N/m$

By applying law of conservation of energy

$l$ = length of incline plane $x=2m$ compression in spring

$\frac{1}{2}k{x}^2=mglsin\theta$
$\frac{1}{2}\times 100\times 2^2=10\times 10\times lsin\theta$
$lsin\theta =2\theta ={30}^{0}l=4m$