Q] Consider the situation shown in the figure. A spring of spring constant 400Nm^-1 is attached at one end to a wedge fixed rigidly with the horizontal part. A 40g mass is released from rest while situated at a height 5cm the curved track. The minimum deformation in the spring is nearly equal to [Take, g=10ms^-2] [AIIMS 2015]
Answers
EXPLANATION.
A spring of spring constant = 400nm⁻¹.
A 40g mass is released from rest while situated at a height 5 cm the curve track.
Minimum deformation in the spring is nearly equal to.
As we know that,
Formula of :
Spring constant = 1/2(kx²).
Potential Energy = mgh.
We can write equation as,
Loss in potential energy = Gain in elastic potential energy.
⇒ mgh = 1/2(kx²).
⇒ 2mgh = kx².
⇒ 2mgh/k = x².
⇒ x = √(2mgh/k).
⇒ x = √(2 x (0.04) x 10 x 5)/400.
⇒ x = √(100 x 0.04)/400.
⇒ x = √(4/400).
⇒ x = √(1/10).
⇒ x = 10 cm ≈ 9.8m.
Explanation:
EXPLANATION.
A spring of spring constant = 400nm⁻¹.
A 40g mass is released from rest while situated at a height 5 cm the curve track.
Minimum deformation in the spring is nearly equal to.
As we know that,
Formula of :
Spring constant = 1/2(kx²).
Potential Energy = mgh.
We can write equation as,
Loss in potential energy = Gain in elastic potential energy.
⇒ mgh = 1/2(kx²).
⇒ 2mgh = kx².
⇒ 2mgh/k = x².
⇒ x = √(2mgh/k).
⇒ x = √(2 x (0.04) x 10 x 5)/400.
⇒ x = √(100 x 0.04)/400.
⇒ x = √(4/400).
⇒ x = √(1/10).
⇒ x = 10 cm ≈ 9.8m.