Question

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

• A. $\sqrt{20}\text{m/s}$
• B. $\sqrt{30}\text{m/s}$
• C. $\sqrt{10}\text{m/s}$
• D. $\sqrt{40}\text{m/s}$

Answer 1

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

Spring force is given by

$F=kx$

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


Given: $x_{max}=2\text{m};M=10\text{kg}$

Now apply energy conservation, between the natural length of spring and its maximum compression state

$U_0+K_0=U_1+K_1$

$Mgh+\frac{1}{2}M{v}^2=\frac{1}{2}k{x}_{max}^2+Mg{h}_1+0$

$\frac{1}{2}M{v}^2=\frac{1}{2}k{x}_{max}^2+Mg(h_1-h)=\frac{1}{2}k{x}_{max}^2+Mg(-x_{max}\sin {30}^{\circ })$

$v=\sqrt{\frac{k{x}_{max}^2}{M}-g{x}_{max}}$

$v=\sqrt{\frac{\left(100\right)\left(2^2\right)}{10}-\left(10\right)\left(2\right)}$

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

Hence, option $\left(a\right)$ is correct.

Why this question ? $W_{ext}=\Delta (U+K)$ If work done by net external force acting on the system is zero, then mechanical energy will be conserved. $K_i+U_i=K_f+U_f$