Question

A Block of Mass 0.5 Kg is placed on top of a light, vertical spring of spring constant 50 N/m, causing the spring to compress some amount. The block is then pushed downwards by a persons hand so that the spring is compressed an additional 15 cm. The block is then released from rest, so that it travels upward and then leave the spring.

A. What is the kinetic energy of the block as it leave the spring?

B. What is the total change in height, from the moment the persons hand releases the block to the highest point it reaches?

Answer 1

Given is :-

Mass of the block m= 0.5 kg

Spring constant K = 50 N/m

Additional compression in the spring x' = 15 cm or 0.15 m

Now,

part - a

Initial compression in the spring

mg= Ka

or

which gives us

therefore the total compression in the spring is

or

Using conservation of energy we get

$KE_{block} = \frac{1}{2}KX^2$

thus

$KE_{block} = \frac{1}{2}(50 N/m)(0.248 m)^2$

which gives us

$\boxed{KE_{block} =1.5376 J}$

part - b

Again applying conservation of energy

or

which gives us

$\boxed{h = 0.313796m}$

or

$\boxed{h = 31.3796cm}$