# Outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring?

## This Post Has 3 Comments

1. Expert says:

explanation:

the masses of the objects and the distance between them

2. Expert says:

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

3. sydmarie02 says:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

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

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.