Question

An ideal massless spring $S$ can be compressed by $1m$ by a force of $100N$ in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane 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. $\sqrt{20}m/s$
• B. $\sqrt{30}m/s$
• C. $\sqrt{10}m/s$
• D. $\sqrt{40}m/s$

Answer 1

The correct option is A $\sqrt{20}m/s$

$F=kx$
$\therefore k=\frac{F}{x}=\frac{100}{1}N/m=100N/m$
Now from energy conservation, between natural length of spring and its maximum compression state.
$\frac{1}{2}m{v}^2+mgh=\frac{1}{2}k{x}_{max}^2$
where $h$ is the height of the block when it just touches the spring from the point where the maximum compression occurs.
$\therefore v=\sqrt{\frac{k{x}_{max}^2}{m}-2gh}$
$=\sqrt{\frac{\left(100\right)(2{)}^2}{10}-\left(2\right)\left(10\right)\left(\frac{2}{2}\right)}$
$=\sqrt{20}m/s$