Question

Consider the situation shown in the figure. A spring of spring constant 400 Nm is attached at one end to a wedge fixed rigidly with the horizontal part. A 40 g mass is released from rest while situated at a height 5cm the curved track. The maximum deformation in the spring is nearly equal to (take $g=10m/s^2$)

• A. 9.8 m
• B. 9.8 cm
• C. .98 m
• D. .009 km

Answer 1

Answer: (B)

9.8 cm

The loss in gravitational potential energy would be equal to the gain in elastic potential energy.

At the point of maximum deformation, all the gravitational potential energy would convert to elastic potential energy.

$⟹mgh=\frac{1}{2}k{x}^2$

$⟹x=\sqrt{\frac{2mgh}{k}}$

$=\sqrt{\frac{2\times 0.04\times 10\times 5}{40}}m$

$\approx 0.098m$

$=9.8cm$