A horizontal spring with spring constant 85 extends outward from a wall just above floor level. A 9.5 box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 before the spring expands and shoots the box back out. How fast was the box going when it hit the spring

0.194 m/s

Explanation:

To solve this question, we're going to be using the law of conservation of energy, which is

Kinetic energy of spring before collision = spring energy of spring after compression

Mathematically, we write it as

½mv² = ½kx², where

k is the spring constant of the spring, 85 N/m

m is the mass of the box sliding towards the spring, 9.5 kg

v is the speed of box just before colliding with the spring and is unknown

x is the compression the spring, 6.5 cm = 0.065 m

You should also know that the kinetic energy of the box just before it collides with the spring converts into the spring energy of the spring when it is fully compressed.

½mv² = ½kx²

Substituting the values we'd stated out, we have

½ * 9.5 * v² = ½ * 85 * 0.065²

4.75 * v² = 42.5 * 0.004225

4.75v² = 0.1796

v² = 0.1796 / 4.75

v² = 0.03781

v = √0.03781

v = 0.194 m/s

6/12=1/2

step-by-step explanation:

answer and explanation;