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

Answer:

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

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$x=v\sqrt{\frac{m}{k}}$

$x=2.5\times \sqrt{\frac{4}{30}}$

$x=0.912\ m$

Answer 2

Answer:

91 cm

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.