Question

A 350-g block is dropped onto a vertical spring with spring constant k =110.0 N/m. The block becomes attached to the spring, and the spring compresses 50.6 cm before momentarily stopping. The speed just before the block hits the spring is 8.4 m/s. The work done by the spring force is -14.1J. The work done by weight is 1.74 J. If the speed at impact is doubled, what is the maximum compression of the spring

Answer 1

The block is dropped from a certain height on the spring. In this way, the block losses potential energy and gains the kinetic energy.

Suppose v is the velocity of the block just before it hits the spring. [This velocity is given as 8.4 m/s]

So, total energy of the spring-block system just before the block strikes the spring = KE of the block = (1/2)mv^2

This energy will convert into the potential energy of spring in the form of its compression.

This is, (1/2)kx^2

So, (1/2)mv^2 = (1/2)kx^2

Now, the speed of impact is doubled. Suppose the new compression of the spring is x1.

So, (1/2)m(2v)^2 = (1/2)kx1^2 = (1/2)m4v^2 = 4*(1/2)kx^2

=> x1 = 2x = 2*50.6 = 101.2 cm = 1.012 m

So, maximum comprerssion of the spring = 1.012 m