# Springs - jpsaos # Springs

A coiled mechanical device that stores elastic potential energy by compression or elongation

Elastic Potential Energy – The energy stored in an object due to a deformation.

## Force Applied to a Spring

x

0

(unstretched position)

F x

1

F=k(x

1

•x

0

-x

0

) is at the original unstretched position of the spring endpoint, x

1

= is the position of the spring after being stretched. Commonly written as

F=kx , x=x

1

-x

0

• x is the amount of compression or elongation from the equilibrium position measured in meters.

•k – spring constant (force constant) – a measure of the resistance of the spring to deformation.

•k is measured in N/m

•The spring scales used in class have a spring constant of 350 N/m (a moderate spring constant)

### Restoring Force

• The resistive force applied by the spring to return to its original shape is called the restoring force (F s

).

• F s

=-kx

. The restoring force is equal in magnitude and opposite in direction to the applied force.

• F=-F s

•The restoring force applied by the spring is called Hooke’s Law

.

Hooke’s Law:

F s

= -kx

F

S

F

## Spring Constant Example

A 0.15 kg mass is attached to a vertical spring and hangs at rest a distance of 4.6 cm below its original position. An additional 0.50 kg mass is then suspended from the first mass and allowed to descend to a new equilibrium position. What is the total extension of the spring?

m

1

=0.15 kg x

1

=4.6 cm x=stretched distance m

2

=0.50 kg k=F

1

/x

1

=(.15kg)(9.8m/s 2 )/.046m=32 N/m x=F/k=(m

1

+m

2

)g/k

=(0.15 kg+0.50 kg)(9.8 m/s 2 )/(32 N/m)=

=0.20 m=20 cm x

1

F

1= m

1 g

F=(m

1

+m

2

)g

F

### (Energy stored in a Spring)

Work is the area under a Force-Displacement graph.

x

W= ½ xF= ½ x(kx)

W= ½ kx 2

The work accomplished in moving a spring from a zero reference to a position x.

Since work is a transfer of energy, then potential energy is gained by a spring elongated from its reference position

PE s

= ½ kx 2

Elastic potential energy gained by a spring.

Work Moving a Spring between Two Locations x

1 x

2

The work (stored energy) moving the spring

From position 1 to 2:

W

1  2

= W

2

-W

1

= ½ kx

2

2 ½ kx

1

2 = ½ k(x

2

2 -x

1

2 ) x

1

=0 for an unstretched spring

W= ½ kx

2

2

From position 2 to 3:

W

2  3

= W

3

-W

2

= ½ kx

3

2 ½ kx

2

2 = ½ k(x

3

2 -x

2

2 ) x

3

The work is moving the spring from position 1 to 2 or position 2 to 3 is NOT :

W

1  2

= ½ k(x

2

-x

1

) 2

W

2  3

= ½ k(x

3

-x

2

) 2

Common mistake/misconception!

Example of Work in Moving a Spring between Two Locations

How much work is required to move a spring with a spring constant of 750 N/m from its unstretched position to 2.0 cm?

W= W

0  1

½ k(x

1

2

=W

1

-W

0

-0)= ½ kx

1

2

= ½ kx

1

2 - ½ kx

0

2 = ½ k(x

1

2 -x

0

2 )=

= ½ (750 N/m)(.02m) 2 =0.15 J

How much work is required to move the same spring an additional 3.0cm?

W=W

1  2

=W

2

-W

1

= ½ kx

2

2 - ½ kx

1

2 = ½ k(x

2

2 -x

1

2 )=

½ (750N/m)[(.05m) 2 -(.02m) 2 ]

=0.79 J

### (Springs)

v

1 m

Time 1 x

1 m v

2 x

1 v

1 x v

2

2

=spring endpoint position at time 1

=mass velocity at time 1

=spring endpoint position at time 2

=mass velocity at time 2

Time 2 x

2

### Energy Equations

ME

1

=ME

2

PE

1

+KE

1

=PE

2

+KE

2

½ kx

1

2 + ½ mv

1

2 = ½ kx

2

2 + ½ mv

2

2

Conservation of Mechanical Energy for Springs

### Example Problem

A 10 kg mass traveling at 1.0 m/s on a frictionless surface compresses an originally undeformed spring with a constant of 800 N/m until it stops. What is the distance that the spring is compressed?

ME

1

=ME

2

PE

1

+KE

1

=PE

2

+KE

2

½ kx

1

2 + ½ mv

1

2 = ½ kx

2

2 + ½ mv

2

2

0+ ½ (10 kg)(1.0 m/s) 2 = ½ (800 N/m)x

2

2 +0 x

2

= 0.11 m = 11 cm