A spring with a constant of $5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $6 kg$ and speed of $2 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

2273 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T07:39:03

Spring will compress by $"2.19 m"$

Kinetic energy of object = Energy stored in spring

$1/2mv^2 = 1/2kx^2$

From above equation

$x = vsqrt(m/k) = "2 m/s" × sqrt("6 kg"/("5 kg/s"^2)) = "2.19 m"$

Answer 2 · 2018-03-10T09:12:23

The compression is $=2.19m$

Mass of the object is $m=6kg$

Speed of the object is $v=2ms^-1$

The kinetic energy of the object is

$KE=1/2mv^2=1/2*6*(2)^2=12J$

This energy will be stored in the spring

$E=1/2kx^2$

The spring constant is $k=5kgs^-2$

Let the compression of the spring be $=xm$

Therefore,

$12=1/2*5*x^2$

$x^2=(24)/5=4.8m^2$

The compression of the spring is

$x=sqrt4.8=2.19m$

A spring with a constant of 5 (kg)/s^25 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 6 kg6 kg and speed of 2 m/s2 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?