Conservation Laws Elastic Energy

Springs When a spring is stretched or compressed a distance x A restoring force, Fs , is created in the spring Acts in a direction to restore the spring to its original length The force is proportional to the displacement x The spring stores elastic potential energy Elastic energy is proportional to the spring constant And is proportional to the square of displacement x

HOOKE’S LAW Fs is the force generated in the spring, measured in Newtons ( N ) k is the spring constant and is related to the springs stiffness, ( N/m ) x is the the distance the spring is stretched or compressed, ( m ) There are many objects that behave like springs: A stretched rubber band, a bent piece of flexible metal, a bow string, and even pendulums. These all have a restoring force and they all obey Hooke’s Law. The negative sign indicates that the restoring force is opposite the applied force that displaced the spring, or spring like object. Displace a spring right and the restoring force acts to the left.

Fs Fg Determining the Spring Constant x m The most common method to find the spring constant is to hang a mass from the spring and lower the mass a distance x to the equilibrium position. The equilibrium position is the location where forces are equal and opposite. m x Fs Fg When the spring/mass system is positioned at equilibrium and subsequently released from rest, it will remain stationary.

Graphing Hooke’s Law m Fs (N) x (m) 2 4 6 8 10 m To stretch a spring a force F must be applied. As the spring stretches a restoring force Fs is created in the spring. The restoring force will be equal and opposite the applied force. The spring constant in this example is k = 10 N/m . 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced zero distance, x = 0 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law Fs F m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced 2 m distance, x = 2 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law Fs F m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced 4 m distance, x = 4 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law Fs F m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced 6 m distance, x = 6 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law Fs F m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced 8 m distance, x = 8 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Graphing Hooke’s Law Fs F m Fs (N) x (m) 2 4 6 8 10 m When a spring is displaced 10 m distance, x = 10 m , the resulting restoring force is 100 80 Fs (N) 60 40 20 2 4 6 8 10 x (m)

Slope of Fs vs. x Fs F m Fs (N) 2 4 6 8 10 m 100 80 Fs (N) 60 The slope of the Hooke’s law graph is the spring constant. 40 20 2 4 6 8 10 x (m)

Example 1 Fs Fg m (A) Determine the spring constant, k . 25 cm A 100 g mass is attached to a spring and is lowered 25 cm until it reaches equilibrium. m (A) Determine the spring constant, k . 25 cm Fs Fg

Example 1 m (B) Determine the energy stored in the spring. 25 cm A 100 g mass is attached to a spring and is lowered 25 cm until it reaches equilibrium. m (B) Determine the energy stored in the spring. 25 cm While this energy can be calculated, is it significant? If the mass is removed from the spring, then the spring will move to restore back to its rest position. Without the mass attached the spring possesses the potential energy calculated above.

Example 1 m This can only occur if x = 0 A 100 g mass is attached to a spring and is lowered 25 cm until it reaches equilibrium. m (C) Determine the energy stored in the spring/mass system. 25 cm If the mass remains attached to the spring, then the spring/mass system will not move when released at equilibrium. As a result, the spring/mass system DOES NOT possess USEFUL potential energy (The potential energy cannot change into kinetic energy). This can only occur if x = 0

Example 1 A 100 g mass is attached to a spring and is lowered 25 cm until it reaches equilibrium. m The original stretch of 25 cm is irrelevant, and is ignored in most potential energy calculations. The equilibrium position is reset as the zero displacement, x = 0 , location for potential energy calculations. 25 cm x = 0

Example 1 m m x = 0 The spring is stretched an additional 10 cm. A 100 g mass is attached to a spring and is lowered 25 cm until it reaches equilibrium. m 10 cm m The spring is stretched an additional 10 cm. (D) Determine the energy stored in the spring/mass system. x = 0

Example 2 A toy gun consists of a spring in a tube that is used to launch a ball horizontally. The spring has a spring constant of 40 N/m and is compressed 20 cm. The ball has a mass of 0.10 kg and is initially at rest. Determine the speed of the ball when it exits the tube. This is a conservation of energy problem. Final Conditions v m x = 0 Initial Conditions x0 m v0 = 0

Example 3 R h0 = R x0 = 0 v0 = 0 R h = 0 x = 40 cm v = 0 A 2.0 kg mass is released from rest and moves down the frictionless surface shown below. The quarter circle section has a radius of 3.0 m. At the end of the horizontal surface the mass collides with a spring compressing the spring 40 cm. Determine the spring constant. R Initial Conditions h0 = R x0 = 0 v0 = 0 R Final Conditions h = 0 x = 40 cm v = 0