Question

A 1.40 kg block is attached to a spring with spring constant 16.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s . What are

Answer 1

A = 0.13 m

Step-by-step explanation:

Given that,

Mass of a block, m = 1.4 kg

Spring constant of the spring, k = 16.5 N/m

While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s or 0.45 m/s

We need to find the amplitude of the subsequent oscillations.

We can use the conservation of energy here. Initially, the kinetic energy of the block is maximum and then it gets converted to the potential energy of the spring.

Mathematically,

`(1)/(2)mv^2=(1)/(2)kA^2`

A is the amplitude of subsequent oscillations.

`A=\sqrt{(mv^2)/(k)} \\\\A=\sqrt{(1.4* (0.45)^2)/(16.5)} \\\\A=0.13\ m`

So, the amplitude of subsequent oscillations is 0.13 m.