Transcript Slide 1

Aim: How can we calculate the
energy of a spring?
HW #33 due tomorrow
Do Now:
An object is thrown straight up. At the maximum
height, it has a potential energy of 300 J. How
much work did it take to do this?
300 J (Work – Energy Relationship)
Convert the mass into kg and
calculate the weight:
distance
mass (g) mass (kg) weight (N) stretched (cm)
50
0.050
0.49
250
0.250
2.45
450
0.450
4.41
650
0.650
6.37
850
0.850
8.33
Plot the data
Force or weight (N)
10
9
8
7
6
5
4
3
2
Your graph should be
a straight line similar
to this one!
1
0
Distance stretched (cm)
Calculate the slope of your graph
Hooke’s Law
F = Force or weight applied (N)
F = kx
k = Spring constant (N/m)
Each spring has a different value
(slinky – low k)
(shocks in car – high k)
*show springs*
x = Distance stretched or compressed
from equilibrium (m)
Robert Hooke
1635-1703
Problems
1. A 4 N weight is hung from a spring causing it
to stretch a distance of 0.16 m. What is the
spring constant?
F = kx
4 N = k(0.16 m)
k = 25 N/m
Next Problem
2. A 5 kg mass is hung from the same spring.
How far does it stretch?
F = kx
mg = kx
(5 kg)(9.8 m/s2) = (25 N/m)x
x = 1.96 m
Elastic Potential Energy
•The energy stored in a spring or any other
elastic item (i.e. – rubberband)
•Represented as PEs
Slinky
Commercail
•PEs = ½kx2
•On a F-x graph:
Area = PEs
F
x
Ace
Ventura –
Slinky
Scene
Problems
1. A spring with a constant of 30 N/m stretches 0.5
m. What is its elastic potential energy?
PEs = ½kx2
PEs = ½(30 N/m)(0.5 m)2
PEs = 3.75 J
Next Problem
2. A spring with a constant of 40 N/m has 12 J of
elastic potential energy. How far has it
stretched?
PEs = ½kx2
12 J = ½(40 N/m)x2
x = 0.77 m
Force or weight (N)
Solve for the elastic potential energy
PEs = area under curve
70
PEs = ½bh
PEs = ½(2 m)(70 N)
PEs = 70 J
2
Distance stretched (m)
Conservation of Energy
An 80 kg girl jumps on a trampoline with 1000 J of
elastic potential energy stored in the springs. How
high with she go?
PEs = PE
PEs = mgh
1000 J = (80 kg)(9.8 m/s2)h
h = 1.3 m
Conservation of Energy
A 100 kg object is connected to a spring with a spring
constant of 5,000 N/m and is compressed a distance of
0.75 m. How fast will the object travel when released?
PEs = KE
½kx2 = ½mv2
kx2 = mv2
(5,000 N/m)(0.75 m)2 = (100 kg)v2
v = 5.3 m/s