Consider the situation shown in figure. A spring of spring constant 400 N / m 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 m / s 2 )

Question

Consider the situation shown in figure. A spring of spring constant 400 N / m 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 m / s 2 ) (A) 9.8 m (B) 9.8 cm (C) .98 m (D) .009 km

Tardigrade Admin · Accepted Answer

Applying the conservation of momentum to the system

m g h = \frac{1}{2} k x^{2}

Given,

m = 0.04 k g, h = 5 m, K = 400 N / m

and

x =

deformation (compression) in the spring

\Rightarrow x = \frac{2 m g h}{k} = \frac{2 \times 0.04 \times 10 \times 5}{400}

= \frac{1}{10} m = 10 c m \approx 9.8 c m

9.8 m

9.8 cm

.98 m

.009 km

Applying the conservation of momentum to the system mgh=21​kx2 Given, m=0.04kg,h=5m,K=400N/m and x= deformation (compression) in the spring ⇒x=k2mgh​​=4002×0.04×10×5​​ =101​m=10cm≈9.8cm