Question

A light spring of length 20cm and force constant 2kg/cm is placed vertically on a table.A small block of mass 1kg falls on it.The length h from the surface of the table at which the balls will have the maximum velocity is

Answer 1

Answer:

The length h from the surface of the table at which the block will have the maximum velocity is 15.1 cm.

Explanation:

Given data:

Length of the light spring, l = 20 cm  

Force constant, k = 2 kg/cm  

Mass of the small block, m = 1 kg

To find: Length h from the surface of the table when velocity is maximum

Let the spring be compressed by a distance of “x” at a certain moment. Therefore, by using the law of conservation of energy at this moment, we can write

mg (h+x) = ½ mv² + ½ kx²

or, ½ mv² = mg (h+x) – ½ kx² …. (i)

[The velocity of the block will be maximum when the kinetic energy is maximum. Also, the velocity keeps on increasing as long its acceleration is increasing]

∴ d[1/2 mv²]/dx = mg – kx = 0

⇒ mg – kx = 0

⇒ x = mg/k …. (i)

Now, to get the height “h” from the surface of the table at which the velocity is maximum, we need to subtract the distance of compression in the spring from the length of the spring, so,

h = l – mg/k ……. [from (i)]

⇒ h = 20 – [1 * 9.8 / 2] = 20 – 4.9 = 15.1 cm