Physics Notes PDF

## Hooke's Law - A stretched spring can also possess stored energy. This is called elastic potential energy *(Ep)*. - Since a stretched spring can do work as it returns to its original position, it must possess stored energy. The same is true for an elastic band or an arrow pulled back in a bow. -...

## Hooke's Law - A stretched spring can also possess stored energy. This is called elastic potential energy *(Ep)*. - Since a stretched spring can do work as it returns to its original position, it must possess stored energy. The same is true for an elastic band or an arrow pulled back in a bow. - Springs can have different amounts of stiffness, as indicated by the spring constant *k*. ## Hooke's Law Lab - A graph of force vs. stretch is a linear relationship. - **What is represented by the slope of this graph?** - *k* = *F*/ *x* = 4N/0.4m = 10N/m ## Hooke's Law - Imagine that you have stretched a spring to a positive displacement. As you hold it there the spring exerts an equal and opposite force in your hand (Newton's Third Law). This force is to the left in a negative direction and attempts to restore the spring to its equilibrium position. This force is called the restoring force. - **Hooke's Law:** *F* = - *kx* - Where: - *F* is the restoring force in Newtons - *k* is the spring constant (N/m) - *x* is the displacement in m - If too much force is applied to a spring, it may become permanently deformed or may even break. Use of excessive force destroys the elasticity of the spring. ## Example: - **What is the elastic force a spring will exert if it has a spring constant of 175 N/m and is stretched 30.0 cm?** - *F* = *Kx* = 175 N/m (0.300m) - *F* = 52.5N ## Example: - **What is the spring constant for a mass of 2.0 kg hanging on a spring stretched 4.0 cm from its rest position?** - *F* = *Kx* - *mg* = *Kx* - *K* = *mg*/ *x* = (2.0kg)(9.81m/s²)/ 0.040m - *K* = 490 N/m ## Energy in a Spring The energy stored in a spring (elastic potential energy) has the ability to do work. - Where, - *Ep* = elastic potential energy *(J)* - *k* = spring constant *(N/m)* - *x* = amount of stretch or compression *(m)* - *Ep* = 1/2 *kx² ## Energy in a Spring - We can look at the work required as the area under a graph of force vs. stretch. ## Energy in a Spring - The area under the graph represents the work done on the spring. - *Area* = 1/2 *bh* - *Area* = 1/2 *Fx* - But *F* = *kx* - So, - *Area* = 1/2 *kx.x = kx²/2 = work ## Example: - A spring with a force constant of 240 N/m has a 0.80 kg mass suspended from it. What is the extension of the spring, and how much elastic potential energy does it have once the mass is suspended? - *F* = *Kx* - *mg* = *Kx* - *x* = *mg*/ *K* =(0.80kg)(9.81m/s²)/ 240 N/m - *x* = 0.0327 m - *Ep* = 1/2 *kx² = (240N/m)(0.0327m)²/ 2 = 0.13J ## Example: - How much work must be done to compress a spring 4.0 cm if the spring constant is 55 N/m? - *W* = *Ep* = kx²/2 = 55N/m (0.040m)²/ 2 = 0.044J ## Example: - A toy gun has its spring compressed 3.0 cm by a 50.0 g projectile. The spring constant is 410 N/m. Calculate the velocity of the projectile if it is launched horizontally. - When spring is compressed → Ep - When projectile is launched → Ek (as it leaves spring) - *Ep* = *Ek* - *Kx²*/4 = *mv²*/4 - *V²* = *Kx²*/ *m* = (410N/m) (0.030m)²/0.0500 kg - *V* = 2.7 m/s ## Simple Harmonic Motion - Repetitive movement back and forth through an equilibrium, or central position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same, and the force responsible for the motion is always directed toward the equilibrium position and is directly proportional to the distance from it. - A mass on a horizontal spring is an example of simple harmonic motion. ## Simple Harmonic Motion - We can use Newton's second law to find an expression for the acceleration of a moving mass on a spring. - *Fnet* = *ma* - *Fnet* = -*kx* - *ma* = -*kx* - *a* = -*kx*/ *m* ## Example: - A 2.0 kg mass on a horizontal spring is stretched 0.30 m from the equilibrium position and released. The spring constant is 65 N/m. - a) What is the initial elastic potential energy of the spring? - *Ep* = kx²/2=(65N/m) (0.30m)² /2= 2.9J *Ep* = 2.9J - b) What maximum speed does the mass reach? - *Vmax* at equil. *Ek* = *Ep* - 1/2 *mu²* = 1/2 *kx² - *V²* = *kx²*/ *m* = (65N/m)(0.30m)²/ 2.0kg - *V* = 1.7m/s - c) At 0.20 m displacement, what is the speed of the mass? - (*Ek*+*Ep*) 0.20m = (*Ep*)0.30m - 1/2 *mV²* + 1/2 *kx²= 1/2 *kx² - *V²* = *Kx²*/ *m* = (65N/m)(0.30m)²/ 2.0kg - 65 (0.20)²/ 2.0 - 2.0 *V²* + 2.6 = 5.85 - *V* = 1.3 m/s - d) What is the maximum acceleration? - *a* = -*Kx*/ *m* = -(65N/m)(0.30m)/2.0kg = -9.8 m/s² - e) What is the acceleration when the displacement is 0.20 m? - *a* = -*Kx*/ *m* = -65N/m) (0.20m)/2.0kg = -6.5m/s² ## Example: - For a spring with a spring constant of 9.0 N/m, calculate the acceleration of an attached 10.0 g mass when the mass is displaced 5.0 cm to the right of its rest position. If the mass is released calculate its velocity: - a) at 2.0 cm to the right of its rest position. - b) at rest position. - c) 5.0 cm to the left of its rest position. - a) (*Ep*) 5.0cm = (*Ep*+*Ek*) 2.0cm - *Kx²*/2 = *Kx²*/2 + 1/2 *mv²* - (9.0N/m) (0.050m)²/2 = (9,0N/m(0.020m)²+ 0.0100kg. *V²*/2 - 0.0225 = 0.0036 + 0.0100*V²* - *V* =1.4m/s - b) *V* = 0 m/s - c) *V* = 0 m/s - a = =-*Kx*/ *m* = -(1.0N/m) (0.050m) / 0.010kg = -45 m/s² ## Conservation of Energy - We can use all three types of energy studied in this unit to solve conservation of energy problems. - When trying to determine what kind of energy an object may possess at different points, ask these questions. - Gravitational Potential: Is the object at a height above the reference point? - Kinetic Energy: Is the object moving? - Elastic Potential Energy: Is there a spring or other elastic medium that has been stretched or compressed? ## Example: - A spring pop-up toy with a mass of 0.020 kg has a spring constant of 220 N/m. If the spring is compressed 0.030 m and the toy is released, calculate the maximum height reached by the toy. - *Eg* = *Ep* - *mgh* = *Kx²*/ 2 - *h* = *Kx²*/ *2mg: = (220N/m) (0,030m)²/ 2(0.020kg) (9.81m/s²) = 0,50 m ## Example: - A ball of mass 0.50 kg is attached to a horizontal spring. The spring is compressed 0.25 m from its equilibrium position and then released. The ball undergoes simple harmonic motion achieving a maximum speed of 1.5 m/s. Determine the spring constant and calculate the speed of the ball when it is halfway to its equilibrium point. - a)*Ep* = *Ek* - *Kx²*/2 = 1/2 *mv²* - *K* = *mv²*/ *x²* = (0.50kg)(1.5m/s)²/ (0.25m)²= 18N/m - b) (*Ep*) 0.25m = (*Ep*+*Ek*)0.125 m - 1/2 *Kx²* = 1/2 *kx²* + 1/2 *mv²* - (18) (0.25m)²=18(0.125m)² + 0.50(V²) - 0.844 = 0.50 V² - *V* = 1.3 m/s ## Example: - A 12 kg ball, starting from rest, rolls down the 10.0 m high track and compresses the spring 2.0 m before stopping. - a) Calculate the spring constant of the spring. - (*Eg*)Top = (*Ep*)bottom - *mgh* = *Kx²*/2 - *K* = 2*mgh*/ *x²* = 2(12kg) (9.81m/s²)(10.0m) / (2.0m)² = 590 N/m - b) Calculate the speed of the ball as it hits the spring - *Eg* = *Ek* - *ngh* = 1/2 *mv²* - *V*= √2*gh* = √2(9.81m/5²)(10.0m) = 14m/s ## Elastic and Inelastic Collisions - Recall our study of momentum and collisions in unit 2. We can learn more about the type collision if we consider conservation of energy principles. ## Types of Collisions - **Elastic Collisions:** Kinetic energy is conserved. (eg. Objects that collide and bounce apart). Momentum is also conserved. - *In elastic collisions, both momentum and kinetic energy are conserved.* - *In the real world purely elastic collisions are hard to achieve because of energy losses due to sound, light, and/or heat due to friction.* ## Types of Collisions - **Inelastic Collisions:** Kinetic energy is not conserved. (eg. Objects that stick together after colliding). Momentum is conserved. ## Example: - A 1200 kg Toyota Corolla moving at 25 m/s[E] collides with a 1300 kg Honda Civic moving at 19 m/s [W]. The Corolla rebounds at 27 m/s. Determine, through calculations, whether the collision is elastic or inelastic. - *MAVA* + *MBVB* = *MAVA'* + *MBVB'* - (1200kg)(25m/s) +(1300kg) (-19mls) = 1200kg(-27mls) + 1300kg *VB'* - *VB'* = 29 m/s [E] - *Ebefore* = 1/2 *1200(25)² + 1/2 *1300 (19)² = 609650J - *Eafter* = 1/2 *1200 (27) 2 + 1/2 *1300 (29) 2 = 984050J - Inelastic ## Example: - A hockey forward with a mass of 95 kg skates in front of the net at 2.3 m/s east. He is met by a 104 kg defenceman skating at 1.2 m/s west. If the players become entangled, - a) Calculate the velocity of the players if they stay together after the impact. - *MAVA* + *MBVB* = (*MA* + *MB*) *V'* - (95kg)(2.3m/s)+(104kg) (-1.2m/s) = (95+104)*V'* - *V'* = 0.47 m/s [E] - b) Determine if the collision is elastic or inelastic. - *Ebefore* = 1/2 *95kg (2.3m/s)²+1/2 *104(1.2)2 = 326.16 J - *Eafter* = 1/2 *199kg (0.47m/s)² = 21J - Inelastic - The document consists of 14 pages of handwritten notes about Hooke's Law and collisions with examples. The document was written on 10/5/2018.