advertisement

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.

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)

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

• F s =-kx • F=-F s . The restoring force is equal in magnitude and opposite in direction to the applied force.

•The restoring force applied by the spring is called

**Hooke’s Law**

.

Hooke’s Law:

**F s = -kx**

F S

**F**

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

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 -**

x 1 =0 for an unstretched spring W= ½ kx 2 2

**½ kx 1 2 = ½ k(x 2 2 -x 1 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 =W 1 -W 0 = ½ kx 1 2 - ½ kx 0 2 = ½ k(x 1 2 -x 0 2 )= ½ k(x 1 2 -0)= ½ kx 1 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

v 1 m Time 1 x 1 m v 2 x 1 =spring endpoint position at time 1 v 1 =mass velocity at time 1 x 2 =spring endpoint position at time 2 v 2 =mass velocity at time 2 Time 2 x 2

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**

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