Unit 7 Notes Energy

Common Types of Energy E g is gravitational potential energy. It is proportional an object’s height above the earth’s surface. – Usually something has 0 gravitational potential energy when it is on the ground. E k is kinetic energy. It is the energy of a moving object. E el is elastic energy. It is the energy stored in a spring or other stretchy object. E chem is chemical energy. This is stored in chemical bonds. We rarely use this in physics. E int is internal energy. This is energy being stored inside an object that is not in a usable form.

Pie Charts The size of the slices shows how the energy is distributed. E int is internal energy. It is energy that is still stored in the system, but not in a usable form (usually thermal energy)

Example: A cart is released at the top of a frictionless incline. At the bottom, it is stopped by a spring. Show a pie chart representation of this scenario (include at least three times).

I chose to show the energy distributions at 4 times: at the top of the ramp, halfway down the ramp, moving on the level surface, and at a complete stop.

Work and Energy W = F ∆x Work is force multiplied by change in position. ***The force and ∆x have to be in the same direction*** Unit: Joules (J) = Newtons x meters (Nm) When a system does work, it gives energy to something else. When work is done on a system, it receives energy. The amount of work done is also the area underneath an F vs ∆x graph

Gravitational Potential Energy F g is basically constant. If I lift up an object to a height h, I do work on it and give it energy W = E g = F g ∆x E g = mgh

Spring Energy In our lab, we found that a spring force is proportional to the change in its length from equilibrium (Hooke’s Law) The area under its F vs ∆x graph is the area of a triangle. Specifically, E el =1/2 k (∆x) 2

Example: A spring has a spring constant of 500N/m. Sketch a graph showing F vs ∆x. How much energy is stored in the spring when it is stretched 0.5m? A = 1/2Bh = ½*0.5m*250N = 62.5J

LOL Charts There will be an L at each time you want to show how the energy in the system is distributed. Between each L, the O shows energy entering or leaving the system. (Usually from external forces).

Example: A cart is released at the top of an incline. At the bottom, it is stopped by a spring. Show an LOL chart representation of this scenario (include at least three times).

Solution: -At the top of the ramp, the cart has gravitational potential energy. -At the bottom of the ramp, some of that energy is now in an unusable form while some is kinetic energy. -When the cart comes to a stop, energy goes into the spring. There is more energy in an unusable form at this point.

Kinetic Energy E k =1/2 m (v) 2 E k is directly proportional to mass and to velocity squared.

Conservation of Energy Energy is never created or destroyed, but it does change forms The initial energy of a system plus any energy added to the system is equal to the final energy of the system plus the energy that leaves the system. E 0 + E input = E f + E out

Steps to Solving 1)Write an equation showing energy at the beginning, at the end, and any energy added to or leaving the system. 2)Show the formulas for these types of energy. 3)Plug numbers into the formula from part 2. 4)Solve for the variable you are looking for.

Example A 2kg ball rolls off a ledge as shown. Solve for its velocity just before it hits the ground. E g + E k = E k mgh + 1/2 m(v 0 ) 2 = 1/2 m(v f ) 2 2kg(9.8N/kg)(2m) + 1/2 (2kg)(2m/s) 2 = 1/2 (2kg)(v f ) 2 V f = 6.57 m/s

A spring shoots a 1kg ball upward. The spring has a spring constant of 300N/m and is compressed 0.1m. How high does the ball fly? E el = E g ½ k (∆x) 2 = mgh 0.5(300N/m)(0.1m) 2 = (1kg)(9.8N/kg) h 1.5J = 9.8N * h h = 0.153m

A 50kg box is sliding to a stop. If it starts moving at 15m/s and there is a coefficient of friction of µ=0.15, how far does the box slide (use energy to solve). ΣF y = F N + F g = 0 F N + 50(-9.8) = 0 490N= F N F f = µ F N F f = 0.15 (490) F f =73.5N E k = E int ½ m (v) 2 = F f ∆x ½ 50 (15) 2 = 73.5 ∆x ∆x=76.5m

Sir Skippy finds the box from the last example and pulls it to the right with a force of 100N. If he pulls it 5m, how much work does he do on the block? Also, solve the box’s velocity. W=F∆x W=100N*5m W=500J W = E int +E k W =F f ∆x + ½ m (v) 2 500J =73.5N (5m) + ½ 50 (v) 2 v = 2.3m/s

Power