Question

A horizontal spring with spring constant 85 N/mN/m extends outward from a wall just above floor level. A 4.5 kgkg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cmcm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring

Answer 1

v = 0.028 m/s

Step-by-step explanation:

In this case you take into account that all the elastic potential of the box, when the spring is compressed is equal to the kinetic energy of the box before it hits the spring. That is:

`K=U`

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

m: mass of the box = 4.5kg

k: spring constant = 85N/m

x: compression of the spring = 6.5cm = 0.0065m

You solve the equation (1) for v:

`v=x\sqrt{(k)/(m)}` (2)

Next, you replace the values of the parameters:

`v=(0.0065m)\sqrt{(85N/m)/(4.5kg)}=0.028(m)/(s)`

The velocity of the box when it hits the spring is 0.028m/s