By: Ryan Carpenter

formulas

-*PE = .5k(x^2)*

*-P = mgh*

1) A spring with a force of 5.2 N/m has a relaxed length of 2.45m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57m. Calculate the elastic potential energy stored in the spring. TOP

x = high – low = 3.57 – 2.45 = 1.12

In this case k is 5.2 so plug it in

PE = .5k(x^2)

PE = .5(5.2)((1.12)^2)

PE = 3.26144

PE = 3.3 J

2) The staples inside a stapler are kept in place by a spring with a relaxed length of .115m. If the spring constant is 51.0N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m? TOP

x = high – low = .150 – .115 = .035

k = 51.0

PE = .5k(x^2)

PE = .5(51.0)((.035)^2)

PE = .03128

PE = 3E-2 J

**3****) A 40.0kg child is in a swing that
is attached to ropes 2.00m long. From
the gravitational potential energy associated with the child relative to the
child’s lowest position under the following conditions**: **TOP**

**a.
****When the ropes are horizontal**

PE = mgh

Swing is 2 meters off ground so height = 2 meters

Gravity = 9.18m/s

PE = (40.0)(9.81)(2)

PE = 784.8

PE = 785J

**b.
****When the ropes make a 30 degree angle with the vertical**

X 2m

Now we have to do some trig. We are just looking for the vertical length so we will solve for h using the angle of 30 degrees. After we find that, we will subtract it from the length of two meters and get the distance the kid traveled up from the bottom.

cos30 = h/2

2cos30 = h

h= 1.73205

2-h = 2- 1.73205 = .26795

now the potential energy

PE = (40.0)(9.81)( .26795)

PE = 105.14358

PE = 105 J

**c.
****At the bottom of the circular arc**

He is at the bottom, therefore 0 relative to the problem