Question

An ideal massless spring $S$ can be compressed $1m$ by a force of $100N$ in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at 30${}^{\circ }$ to the horizontal. A $10kg$ block $M$ is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by $2m$. If $g=10m{s}^{-2}$, what is the speed of mass just before it touches the spring?

• A. $\sqrt{20}$ms${}^{-1}$
• B. $\sqrt{30}$ms${}^{-1}$
• C. $\sqrt{10}$ms${}^{-1}$
• D. $\sqrt{40}$ms${}^{-1}$

Answer 1

The correct option is A $\sqrt{20}$ms${}^{-1}$

$F=kx$

$\therefore k=\frac{F}{x}=\frac{100}{1}=100N/m$

$E_i=E_f$

$\frac{1}{2}{mv}^2=\frac{1}{2}{kx}^2-mgl\sin \left(\theta \right)$

$\therefore \frac{1}{2}\times 10\times v^2=\frac{1}{2}\times 100\times (2{)}^2-(10\left)\right(10\left)\right(2\sin {30}^{\circ })$

$\text{Solving we get}$

$v=\sqrt{20}m/s$