Question

A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has a spring constant of 30Nm. What is the maximum compression of the spring after the collision?

Answer 1

Step-by-step explanation:

Given

mas of block
`m=4\ kg`

speed of block
`v=2.5\ m/s`

spring constant
`k=30\ N-m`

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

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

`x=v\sqrt{(m)/(k)}`

`x=2.5* \sqrt{(4)/(30)}`

`x=0.912\ m`

Answer 2

91 cm

Step-by-step explanation:

mass, m = 4 kg

velocity, v = 2.5 m/s

spring constant, K = 30 N/m

Let the compression in spring is y.

The kinetic energy of the block is converted into elastic potential energy of the spring.

1/2 mv² = 1/2 Ky²

4 x 2.5 x 2.5 = 30 x y²

y = 0.91 m = 91 cm

Thus, the maximum compression in the spring is 91 cm.