Elastic P.E. - work stored in a deformed spring
a) General Equation PEs = Fx
F = Average Force needed to deform spring
x = distance deformed (stretch or compress)
Noble and Greenough School High School
b) Hooke's Law - force needed to deform an ideal spring, a given amount is directly proportional to its displacement (deformation)
Ex) Ideal Spring
| F
(Force to Deform) (N) |
X
(distance deformed) (m) |
| 2.0 | .30 |
| 4.0 | .60 |
| 6.0 |
.90 |
| 8.0 |
1.2 |
slope = F/x
x = stretch
c) Spring constant for ideal spring
- quantifies stiffness of a spring
| K | = | F | (N) |
| X | (m) |
What is the spring constant of the spring that produced the data? above?
| F
(Force to Deform) (N) |
X
(distance deformed) (m) |
| 2 | .30 |
| 4 | .60 |
| 6 | .90 |
| 8 | 1.2 |
| k | = | 20. N |
| .30 m |
k = 6.7 N/m
Which is the stiffer spring?
Spring A is the stiffer spring.
More
force is required
for the same stretch
Ex) Plots and K
Pick any point on the line
| K | = | F | (N) |
| X | (m) |
| K | = | 20. N |
| .40 m |
K = 50. N/m
d) Elastic P.E. = area of triangle under F vs x graph
PEs = (½)basexheight
e) PEs in terms of k
1. P.E.s = Fx
Average Force: F = F/2
2. PE = (F/2)x
substitute F = Kx (On reference)
3. P.E.s = (1/2)(Kx)x
| P.E.s = 1/2(kx2) |
(On reference)
Ex 1) A force of 12. N stretches a spring and makes it .15m longer
a) What the spring constant (k) of this spring?
Ex 1) A force of 12. N stretches a spring and makes it .15m longer
a) What's the spring constant (k) of this spring?
k = F/x = 12. N/.15 m = 80. N/m
b) What's the P.E. of this spring?
PEs = (½)Kx2
= 1/2(80. N/m)(.15 m)2
PEs = .90 Joules
Ex) As shown in the diagram below, a 0.50-meter-long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight.
If 15. joules of energy are stored in the stretched spring, what is the value of the spring constant
If 15. joules of energy are stored in the stretched spring, what is the value of the spring constant
| P.E.s = 1/2(kx2) |
15. J = (½)K(.50 m)2
15. J = .50 k(.25)
= 120 N/m