# 1. An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by a distance d.if the same object is attached to the same vertical spring but permitted to fall freely, through what distance does it stretches the spring.2.A block of 2 kg is dropped from a height of 40 cm on a spring of force constant 1960 N/m what will be the maximum distance x to which the spring is compressed.

1.

When the mass is slowly lowered, the spring will extend till the gravitational force on the mass is balanced by the restoring force of the spring. Such that, mg = kd => d = mg/k

When the mass is allowed to fall freely it will attain a kinetic energy which will put the mass in an oscillatory about the mean extension of the spring which is also ‘d’.

2.

Initial PE of the block is = mgh = (2)(10)(0.4) = 8 J  [g = 10 m/s2]

When it hits the spring the compression of the spring is such that the mechanical energy of the block is equal to the potential energy stored in the spring.

So, ½ kx2 = 8

=> x2 = (2)(8)/(1960) = 0.008

=> x = 0.09 m = 9 cm

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