Transcript Springs and Hooke`s Law

Springs and Hooke’s Law
Physics 11
Springs
A mass-spring system is given below.
As mass is added to the end of the
spring, what happens to the spring?
WHY???
Answer
 Gravitational force (Fg) or the weight of the
mass pulls the spring down (stretches
spring). This creates potential ELASTIC
energy (energy due to the shape of the
spring).
 The more mass on the end of the spring,
the farther it goes down and the more
potential energy it has.
 The force from the spring is equal to the
force of gravity.
Springs – What happens when you
add more weight?
Fspring
x
Fg  1m g
x
Fg  2m g
x
x
Fg  3m g
Fg  mg
 Fg  Fspring
Springs
 2 times the mass results in a 2 times of
the displacement from the equilibrium
point…
 3 time the mass… 3 times the
displacement…
 Fg  Fspring
Fspring  kx
 mg  kx
 2m g  k 2 x 
What kind of energy is this?
 Potential Energy
 Elastic Potential Energy to be exact!
Compression
 Springs can also compress. If you
compress a spring it can gain
potential energy as well. When you
let go, the spring transforms the
potential elastic energy into another
type of energy (kinetic in the case of
the push toys).
What else besides springs has
elastic potential energy?
 Diving boards
 Bows (bow and arrows)
 Bungee cord
Elastic Energy – Summary Slide
 Ee = The potential energy that is stored in
elastic/stretchy things like: elastics,
springs, diving boards, bungee cords, bows
(bow and arrows), etc.
 Elastic potential energy is due to the shape
of the elastic or spring- either compressed
or stretched.
Elastic or Spring Force Summary
 The force that is used to create the
compression or stretch in the
spring/elastic.
Fspring  kx
 This equation will be explained soon 
Restoring Force Summary
 The restoring force is the force that is
needed to put the spring back to
equilibrium. It is in the opposite direction
of the force that compressed or stretched
the spring to store the energy originally.
 Example: If you stretch a spring by 0.5m
and you had to use 150N of force, the
restoring force is -150N.
Hooke’s Law
 The restoring force
is opposite to the
applied force.
(negative sign)
 Gravity applied in the
negative direction, the
restoring force is in the
positive direction
Fspring  kx
Hooke’s Law (summary slide)
Fspring  kx
Fspring: Applied force to stretch/compress spring
x : displacement of the spring from the
equilibrium position (units: m)
k: the spring constant (units: N/m)
The spring constant is unique to the spring
(similar to coefficient of friction). A large
spring or coil has a large k value.
Example
 An archery bow requires a force of
133N to hold an arrow at “full draw”
(pulled back 71cm). Assuming that
the bow obeys Hooke’s Law, what is
its spring constant?




F = kx
(Hooke’s Law)
133 = k(0.71)
(sub in values)
k = 133/0.71
(rearrange)
k = 187.32 N/m  190 N/m
Practice Problems
 Textbook
 Page 258
 35-37
 http://www.youtube.com/watch?v=y
XnbvZx9iWs
Elastic Potential Energy of a Spring
(summary)
 Formula: Ee = ½ kx2
 k is the spring constant
 x is the displacement from
equilibrium position
 Units: Joules (J)
Example:
 A spring with spring constant 75 N/m
is resting on a table.
 A) If the spring is compressed a
distance of 28cm, what is the
increase in its potential energy?
 B) What force must be applied to
hold the spring in this position?
Answer:






A) Ee = ½ kx2
Ee = ½ (75)(0.28)2
Ee = 2.9 J
B) F = kx
F= 75(0.28)
F = 21 N
Practice Problems
 Page 261, questions 38, 39, 40
 Page 261 (Section Review)
 1, 2, 3, 4, 7
Conservation of Energy
 Remember: Energy cannot be created or
destroyed.
 Using the same equation as before, Ei = Ef,
now we can add another type of energy in:
 Eg+Ek+Ee (initial)= Eg+Ek+Ee (final)
 In presence of friction:
 Eg+Ek+Ee (initial)= Eg+Ek+Ee (final)+ Q
Quick Lab – Spring Constant
Conservation of Energy with a
Spring
 Ex. 1: A 4.0 kg block slides across a
frictionless table with a velocity of
5.0m/s into a spring with a stiffness
of 2500 N/m. How far does the
spring compress?
Answer
 X = 0.20m
Example 2:
 A 70. kg person bungee steps off a
50.m bridge with his ankles attached
to a 15m long bungee cord. Assume
the person stops at the edge of the
water and he is 2.0m tall, what is the
force constant of the bungee cord?
Practice Problems
 Textbook
 Page 261
 38-40
 Section review (p 261)
 1-10