Question

3. Consider the situation shown in 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
5 cm the curved track. The minimum deformation in
the spring is nearly equal to (Take, g=10 ms)
(a) 9.8 m
(b) 9.8 cm
(c) 0.98 m
(d) 0.009 km​

Answer 1

Answer:

ANSWER

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=

2

1

kx

2

⟹x=

k

2mgh

=

40

2×0.04×10×5

m

≈0.098m

=9.8cm