- #1

chubbyorphan

- 45

- 0

## Homework Statement

Hi forum. I have a spring type question for anyone willing to help!

k = 225N/m

m = 1.5kg

x = 12.0cm

Friction is not a variable

A spring (with a force constant of 225N/m) is resting on a horizontal(frictionless) surface and is mounted on a wall. A 1.5kg box is pushed against the spring compressing it 12.0cm from equilibrium. When released, the spring pushes the box.

a)how much force must be exerted on the spring to compress it 12.0 cm?

b)how much work is done on the spring to compress it 12.0 cm?

c)how much elastic potential energy is stored in the spring it is compressed?

d) Once released, what maximum speed will the box attain?

**I think I got parts a)b)c) if someone wanted to check it anyway though that'd be really cool, but I definitely need help with part d)

## Homework Equations

a)F = kx

b)W = (1/2)kx^2

## The Attempt at a Solution

here are my answers:

a)F = 27 N

b)E¬e = 1.62 J

c) same as b)

d)We know that maximum speed occurs when Ek is at a maximum and therefore Ee is at a minimum. This point occurs where the spring naturally rests. Therefore, since the spring is compressed 0.120 meters, this point occurs 0.120 meters away upon being released.

Einitial = Efinal

Ee initial = Ek final

1.62 J = (1/2)mv^2

1.62 J = (1/2)(1.5kg)v^2

1.62 J = (0.75kg)v^2

v^2 = (1.62 J)/(0.75kg)

v^2 = (1.62 J)/(0.75kg)

v^2 = 2.16m^2/s^2

v = 1.469693846

After rounding:

v = 1.5m/s

Therefore, the speed of the 1.5kg box as it passes the point 12.0cm or 0.120m from its point of compression will be 1.5m/s

**Is this right^?

Also.. If I wanted to show:

Ee initial +Ek initial = Ee final + Ek final

is this right:

[(1/2)(225N/m)(0.120m)^2][(1/2)(1.5kg)(0)^2] = [(1/2)(225N/m)(0)^2][(1/2)(1.5kg)(v)^2]

Thanks so much in advance to whoever can help me!