Question

A block of mass 2 kg is released from the top of an inclined smooth surface as shown in the figure. If spring constant of the spring is 150 N/m and the block comes to rest after compressing the spring by 1 m, then the distance travelled by the block before it comes to rest is

• A. 6.5 m
• B. 7.5 m
• C. 2.5 m
• D. 5 m

Answer 1

The correct option is B 7.5 m


Loss in gravitational potential energy of the block, U = mgh
Gain in elastic potential energy of the spring, $E=\frac{1}{2}k{x}^2$ (where k = spring constant, x = compression)
Applying conservation of energy:
Loss in gravitational potential energy = Gain in elastic potential energy of the spring
$\Rightarrow mgh=\frac{1}{2}k{x}^2\Rightarrow mg\frac{s}{2}=\frac{1}{2}\times 150\times \left(1{)}^2\right(\because k=150N/m,x=1m,h=s\times sin{30}^{\circ })\Rightarrow \frac{2\times 10\times s}{2}=\frac{150}{2}\Rightarrow s=7.5m$
Hence, the correct answer is option (b).