IIT JEE , AIEEE Forum - Physics , Mechanics

simone

pls pls pls help!

1. A light spring is hung vertically from a fixed support and a heavy mass m is attached to a lower end. the mass is then slowly lowered to its equilibrium position. this stretches the spring by an amount d. If the same body is permitted to fall instead, through what distance does it stretch the string?
ANS. OPTIONS:
(a) d
(b) 1.5 d
(c) 2d
(d) 3d

imagine_me2

try this:

for lowering slowly use force equation mg=kd ,=>k=mg/d-------1

for fall use conservation of energy 1/2kx^2=mgx-------2

now put the value of k form eq.1 in eq2

solve, and may be you will get x=2d.

cvramana

When the mass is slowly lowered, the mass does not gain velocity. Therefore

at equilibrium balancing the forces we get

k d = m g

d = m g / k ------------------- (1)

If the mass is let fall freely, the mass gains and looses kinetic energy.

work done by the gravity + work done by the spring = change in kinetic energy.

mg h - 0.5 k h² = 0

h = 2 mg / k = 2 d

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