Question

A ring of mass 1 Kg can slide on a smooth vertical rod.The ring is connected to a spring of spring constant 40N/m and natural length of spring is 2 m.The ring is taken at a height of 1.5m and released.The velocity of ring when it reaches the ground is equal to

Answer 1

Answer:

The velocity is $4\sqrt{7}$ m/s

Explanation:

When the ring is at position P, the total elongation of the spring

$l=\sqrt{2^2+1.5^2}$

$\implies l=\sqrt{6.25}$

$\implies l=2.5$ m

Since the natural length of the spring is 0.2 m, the spring has been stretched by 2.5-0.2 = 2.3 m

When the ring is at point C, the elongation of the spring

$l'=2$ m

Total stretch of the spring = 2 - 0.2 = 1.8 m

We know that potential energy of a spring is

$U=\frac{1}{2}kx^2$

Work done by spring = change in potential energy

Work done by all forces on the ring = change in Kinetic Energy

$\frac{1}{2}mv^2=W_s+mgh$

$\frac{1}{2}\times 1\times v^2=\frac{1}{2}\times 40\times (2.3^2-1.8^2)+1\times 10\times 1.5$

$\implies v^2=40\times 2.05+15\times 2$

$\implies v^2=82+30$

$\implies v^2=112$

$\implies v=\sqrt{112}$ m/s

$\implies v=4\sqrt{7}$ m/s

Hope this answer is helpful.