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 ${30}^o$ 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. $\sqrt{20}m{s}^{-1}$
• B. $\sqrt{30}m{s}^{-1}$
• C. $\sqrt{10}m{s}^{-1}$
• D. $\sqrt{40}m{s}^{-1}$

Answer 1

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

$F=kx$

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

From energy conservation between points

1) when spring is compressed and

2) when it is at its natural length

$E_i=E_f$

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+mgh$

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2-mgh$

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

Solving we get,

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