Hooke’s Law - Linville Hooke's Law AP.pdf · • Hooke's Law gives the force acting on a spring or other elastic material when it has been stretched or compressed • The direction

Hooke’s

Law

• Hooke's Law gives the force acting on a

spring or other elastic material when it has

been stretched or compressed

• The direction of the force is opposite to the

direction of stretch

Weight of object stretches spring

Restoring force acts up

force

k spring constant (N/m)

of spring

(change in length)

s

s

F k x

F restoring

x displacement

Hooke’s Law (AP)

• Fs = restoring force

• x = displacement

sF k x

• F is proportional to x

sF k x

Is the equation of a line

Example

• A 400 g mass is hung form the lower end

of a spring. The spring stretches 0.200 m.

Calculate k.

Solution

m/N.k

m.

kg/N.kg.k

x

gm

x

Fk

xkF

s

s

619

2000

8194000

Example

• A 150 g mass is attached to one end of a

horizontal spring (k = 44.3 N/m) and the

spring is stretched 0.104 m.

a) Determine the maximum acceleration

when the mass is released.

x

Solution

2

44.3 / 0.104

0.150

30.7 / left

s netF F

k x m a

k xa

m

N m ma

kg

a m s

Apply Newton’s

2nd Law

b) Determine the acceleration of the object

when it is 0.055 m from the equilibrium

position

2

44.3 / 0.055

0.150

16.2 / left

s netF F

k x m a

k xa

m

N m ma

kg

a m s

• The magnitude of the acceleration is

constantly changing since the restoring

force changes as x changes

• Non-uniform accelerated motion means

you can’t use the kinematics equations

from unit 1

Example

• A spring is connected to a box and its k

determined. The spring is then cut in half

and connected to the box as shown.

Describe the value of k for the two springs

Solution

• It's 2k, since each spring now only extends

half the displacement of the original spring

when subjected to the same force.

More than One Spring

• Two springs are

connected vertically in

series and a mass is

suspended from the

lower end. Derive an

expression for the

effective spring

constant.

More than One Spring

• F1 = -k1x1

• F2 = -k2x2

• The forces on each

spring are equal. (ignore

the spring masses)

• -k1x1 = -k2x2

More than One Spring

• -k1x1 = -k2x2

• F = -keff(x1 + x2)

• -keff(x1 + x2) = -k1x1 + -k2x2

2 21

1

k xx

k

More than One Spring

𝑘𝑒𝑓𝑓𝑘2𝑥2𝑘1

+ 𝑥2 = 𝑘2𝑥2

Divide by x2

• 𝑘𝑒𝑓𝑓𝑘2

𝑘1+ 1 = 𝑘2

More than One Spring

• 𝑘𝑒𝑓𝑓𝑘2

𝑘1+ 1 = 𝑘2

• 𝑘𝑒𝑓𝑓𝑘2+𝑘1

𝑘1= 𝑘2

• 𝑘𝑒𝑓𝑓 =𝑘2𝑘1

𝑘2+𝑘1

1 2

1 1 1

effk k k

More than One Spring

• Two springs are

connected as shown and

a force is applied. What

is the spring constant of

the system? (connected

in parallel)

More than One Spring

• The value of x is the

same for both springs

• F = -k1x + -k2x

• F = -(k1 + k2)x

• keff = k1 + k2

Example • Two identical unstrained springs (k = 15

N/m) are connected to an 1.0 kg object as

shown. If the object is moved 4.0 cm to the

left and released, what is the acceleration

magnitude of the system?

Spring Potential Energy

• the energy stored

in a spring

depends on the

displacement

• the work done to stretch the spring = area

under graph

2

2

1

2

1

2

1

2

1

kxE

xkxE

FxE

abEwork

AP

• Us = ½ kx2

Practice

• P 301: 1, 2, 3, 4, 5