I. Energy & Work

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A radiometer shows that light is energy by converting light energy into heat energy and then into movement (kinetic energy)

I. Energy – ______________________________ - Vector? Scalar?

II. Work

A. Defined – _____________________________

Challenge Question: Two people exert a 10 Newton force on a block for 2 meters. One person exerts her force at a 50-degree angle. (frictionless)

B. Equation: ______________________

$\theta$ - the angle between the ______ and ___________________

C. Positive and Negative Work

Work is negative when ________________________________

Work is positive when _________________________________

Work is 0 when ______________________________________

Examples of negative work ______________________________

Ex) A 5.0 N force pushes a box 3.0 m North and then 4.0 m east. Find the work done.

Teacher CER Key: Module 1

Evidence: Energy is the capacity to do work. Work ($W$) is the transfer of energy ($W = Fd\cos\theta$).

Claim: Energy and Work are Scalar quantities measured in Joules ($J$).

Reasoning: Scalar quantities have magnitude but no direction. In the North/East example: $W_{total} = (5.0 \, N)(3.0 \, m) + (5.0 \, N)(4.0 \, m) = 15.0 \, J + 20.0 \, J = 35.0 \, J$.

II. Work Case Studies

W = Fd

Video Instruction

Ex 1) Object pushed at an angle, f≠0, Mass m, Force F, μ, Fn

Find the Net Work done on the mower in terms of the variables given

Ex 2) Find the minimum work needed to push a 950 kg car 810 m up a 9.0 deg incline,

W ork is minimal when ______________________________________

a) no friction

b) friction μk = 0.25

Ex 3) Textbook p. 162) 8

Instructional Video

330 kg piano, slides down 3.6 m, 28° incline / uk = .40. Man prevents the piano from accelerating down the incline

(a) Force exerted by the man?

b) Work done by a man on the piano?

c) Work done by friction?

d) Work done by the force of gravity?

(e) Net work done on the piano?

D. Finding Work using a F vs. d plot

Instructional Video

Ex 4)

W = _________________

Ex 5) p. 162) 14 a) Sketch the plot described below

b) Find the work

Teacher CER Key: Module 2

Ex 2 Claim: $W_{min} = 1.18 \times 10^6 \, J$. $W_{friction} = 1.86 \times 10^6 \, J$.

Ex 2 Reasoning: $W = mgh = (950.0 \, kg)(9.8 \, m/s^2)(810.0 \, m \sin 9.0^\circ) = 1,179,577.0 \, J$. With friction: $W = \Delta PE + f_k d = 1.18 \, MJ + \mu (mg\cos\theta)d = 1.86 \, MJ$.

Ex 4 Claim: $Work = 15.0 \, J$.

Ex 4 Reasoning: Area of the triangle under the $F$ vs $d$ curve: $0.5 \cdot Base \cdot Height = 0.5 \cdot (6.0 \, m) \cdot (5.0 \, N) = 15.0 \, J$.

III. Potential Energy

Potential Energy / Work - Video Instruction

Potential Energy – Stored energy due to _________________ or ______________

Gravitational PE

Equation - _______________ or ______________

W = ΔPE = mgh

W is + when ___________

W is - when ___________

Ex 1) A 1.60 m person lifts a 2.10 kg book from the ground to 2.20 m above the ground. a) How much work was done lifting the book?

Elastic PE

Equation: PE = Fx F – Average Force x – stretch/compression

Ideal Spring – F and stretch ___________________

Spring Constant k = F/x

Ex 2) A spring with k = 53 N/m hangs vertically on a ruler. The end of the spring is next to the 15 cm mark on the ruler. If a 2.5 kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

PE = [F/2]x = 1/2kx²

Ex) Plot of PE vs x?

Elastic P.E./Plot

PE = ___________ of the triangle under F vs x graph

Ex 3) Find the Elastic PE when the spring is stretched 6 m

Ex 4) If 15. Joules of energy are stored in the stretched spring, what is the value of the spring constant?

Ex 5) A 1200 kg car rolling on a horizontal surface has a v = 65 km/hr when it strikes a horizontal coiled spring and is brought to rest at a distance of 2.2 m. What is the spring constant of the spring?

Teacher CER Key: Module 3

Ex 1 Claim: $Work = 45.3 \, J$.

Ex 1 Reasoning: $W = mgh = (2.10 \, kg)(9.8 \, m/s^2)(2.20 \, m) = 45.3 \, J$.

Ex 4 Claim: $k = 30.0 \, N/m$.

Ex 4 Reasoning: $PE_s = 0.5kx^2 \implies 15.0 \, J = 0.5 \cdot k \cdot (1.0 \, m)^2 \implies k = 30.0 \, N/m$.

IV. Kinetic Energy & Power

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Kinetic Energy / Power - Video Instruction

I. Kinetic Energy

I. Kinetic Energy – Energy of ______________

A. KE = Work to stop or Work to get an object from rest to a certain speed

Derive KE Equation

KE = W = Fd Where Vf = 0

KE = (ma)d

Equation - _____________________

B. KE/m ____________ relationship KE/v ________________________ relationship

Plot shape

C. When Work is done on an object KE is changed:

W = ΔKE

When is the Work done on an object Negative? _______________

When is the Work done on an object Positive? _______________

D. Scalar or Vector?

Ex 1) A softball having a mass of .25 kg is pitched at 95 km/hr. When it reaches the plate, the ball slows down by 10%. Neglecting gravity, estimate the average force of friction if the distance to the plate = 15 m

Ex 2) baseball (m = 140 g) traveling 32 m/s caught by a glove and moving the glove back 25 cm. Average F of the ball on the glove?

Power

Power - Video Instruction

II. Power - the rate of doing ______________ or expending _______________

A. Equation P = W/t = __________ = _______________

B. Scalar or Vector?

C. Units _______________

Ex 3) 1440 kg car accelerates from rest to 95 km/hr in 7.4 sec. What average power is used?

Ex 4) A 95 kg football player runs upstairs in 66 s. The stairs are 140 m long at an angle of 32° (constant V)

Her power output vertically is

Ex 5) 1200 kg car has a max. power output of 120 hp. How steep a hill can it climb at a constant speed of 75 km/hr if the frictional force is 650 N?

1 HorsePower = 746 Watts

Teacher CER Key: Module 4

Ex 2 Claim: $Force = 286.7 \, N$.

Ex 2 Reasoning: $W = \Delta KE \implies Fd = 0.5mv^2 \implies F(0.25 \, m) = 0.5(0.140 \, kg)(32.0 \, m/s)^2 = 71.68 \, J \implies F = 286.7 \, N$.

Ex 4 Claim: $Power = 1047.8 \, Watts$.

Ex 4 Reasoning: $P = mgh/t = (95.0 \, kg)(9.8 \, m/s^2)(140.0 \, m \sin 32.0^\circ) / 66.0 \, s = 1047.8 \, W$.

Ex 5 Claim: $Angle = 10.5^\circ$.

Ex 5 Reasoning: $120 \, hp = 89,520 \, W$. $v = 75 \, km/hr = 20.83 \, m/s$. $P = F_{total} \cdot v \implies F_{total} = 89520 / 20.83 = 4297.6 \, N$. This total force overcomes gravity and friction: $F_{total} = mg\sin\theta + f_k \implies 4297.6 = (1200)(9.8)\sin\theta + 650 \implies \sin\theta = 0.3101 \implies \theta = 18.06^\circ$.

V. Conservation of Energy

pdf Resource Review - Instructional Video

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I. Law of Conservation of Energy - unless ________________is done on or by a system, its total mechanical energy does not change.

Ex 1) Fill in the missing energies

Ignore friction

Ex 2) a) If friction is negligible when the RED ball is released, how high will it rise?

A B C D Cannot be determined

b) If friction is negligible when the BLUE ball is released, how high will it rise?

A B C D Cannot be determined

If KE at A is 0 J, find the KE and PE at each point (frictionless)

Ex 3) Fill out the chart below

	KE	PE
A	0
B
C
D
E
F
G

Ex 4) What physics law explains why the cart will not reach point H and I?

Ex 5) If PE at A = 10. J, KE = 0, and PE at C = 3.0 J, find the PE and KE at all other points

Ex 6) 1.0 kg pendulum swings to a height of .20 m above its lowest point. K.E. of the pendulum at the lowest point?

Ex 7) 5.00-kg cart at the foot of a hill 10.0 m high. For the cart to reach the top of the hill, what is the minimum KE of the cart in the position shown?

Ex 8) Find the height of the pendulum in terms of θ and L (pendulum length)

Teacher CER Key: Module 5

Ex 1 Claim: $TME = 100.0 \, J$ at all points.

Ex 1 Reasoning: $\Sigma E = PE + KE$. Frictionless systems conserve mechanical energy. If $PE = 100.0 \, J$ and $KE = 0.0 \, J$ at start, total is always $100.0 \, J$.

Ex 7 Claim: $KE_{min} = 490.0 \, J$.

Ex 7 Reasoning: $KE_{start} = PE_{peak} = mgh = (5.00 \, kg)(9.8 \, m/s^2)(10.0 \, m) = 490.0 \, J$.

VI. Advanced COE & Work

pdf Resource Instructional Video

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Relationship: $W = \Delta KE + \Delta PE + W_f$

Energy Problems

1. Neglecting air resistance, how high will a 0.325-kilogram rock go if thrown straight up by someone who does 115 J of work on it?

2. A hammerhead with a mass of 2.0 kilogram is allowed to fall onto a nail from a height of 0.40 meters. What is the maximum amount of work it could do on the nail?

3. A 1,200-kilogram car traveling at 110 kilometers per hour (30.55 m/s) comes to a stop.

a. How much work (negative) was done on the car? Don't solve, just show the equation.

b. Where does the car's kinetic energy go?

4. A baseball (m = 0.140 kilograms) traveling 35 meters per second moves a fielder's glove backward 0.25 meters when the ball is caught and brought to rest. What is the average force exerted by the ball on the glove?

5. In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy. With what minimum speed must the athlete leave the ground to lift her center of mass 2.10 meters across the bar at 0.70 meters per second?

Conservation of Energy Problems

Video Instruction

1.

a. A 5.0-kg block starts at a speed of 10. m/s at the top of a "roller-coaster," as shown above. What is its KE at the top?

b. What is its GPE at the top (relative to the ground)?

c. What is the block's total energy at the top?

2. If the track is entirely frictionless and there is no air resistance, the block will not lose any energy until it reaches the level surface where μ=0.5 and the frictional force "takes away" energy.

a. Find the GPE at point A and then calculate how much of the block's total energy must be in the form of KE. Also, find the speed at point A.

b. Repeat the above for points B, C, and D.

3. After the block exits the loop-the-loop, it encounters a horizontal, rough surface (μ = 0.5) and comes to a stop.

a. How much total energy does the block have just as it exits the loop-the-loop?

b. How much energy does the frictional force "take away" in bringing the block to a stop?

c. How large is the frictional force acting on the block?

d. How far does the block slide before coming to a stop?

Conservation of Energy Problems with Work

Instructional Video (1 - 2)

1. A 20. kg block, originally at rest, is pulled across a rough surface (μ = 0.10) by a 100. N force angled at 36.87° for 20. m. Find the work done by each of the following forces.

a. Find the work done by the force of friction

b. Find the work done by Force 100. N.

c. Find the total work done by all of the above forces. _____________

d. Find the speed of the block the instant it is 20. meters to the right of its starting position.

2. A 20.-kg block slides across a frictionless floor and then slides up 10. meter along a rough 36.87° incline (μ =0.20) until it comes to a brief stop. What is the initial speed of the block prior to sliding up the incline?

3. A 5.0 kg box is pulled up a rough (μ = 0.10) incline by a force of 50. N for a distance of 20. m. If the speed of the box is 5.0 m/s initially, what is its speed the instant it has traveled the 20. -m up the incline?

Equation________________________________

4. A 10.-kg box is released from rest and slides down an 800. cm tall incline and then compresses a spring (k = 50. N/m). What is the maximum spring compression before the box is pushed away?

Note: all surfaces are frictionless except the crosshatched surface where d = 10. m and has a coefficient of friction of 0.10.

5. A 10.-kg box is held against a spring, which is compressed 2.0 m as shown. If the box is moving at 10. m/s when it reaches the top of the incline, what is the spring constant (k)?

Note: all surfaces are frictionless, and the height (H) of the incline is 6.0 meters.

6. The system above is released from rest. The instance the suspended mass has fallen 8.0 meters, the blocks have speeds of 8.0 m/s. ($m_{susp} = 6.0 \, kg$, $m_{table} = 4.0 \, kg$).

a. How much GPE did the suspended mass "lose" the instant the suspended mass has fallen 8 meters?

b. Where did the GPE "lost" go? Note: disregard air resistance

c. What is the KE of each block, the instant the suspended mass has fallen 8 meters?

d. How much energy did the frictional force "take away" from the system (neg. work)?

e. How large is the frictional force?

f. What is the coefficient of friction?

Teacher CER Key: Module 6

Energy Prob 1 Claim: Height $= 36.1 \, m$.

Energy Prob 1 Reasoning: $W = mgh \implies 115.0 \, J = (0.325 \, kg)(9.8 \, m/s^2)h \implies h = 36.1 \, m$.

COE Prob 1c Claim: $E_{total} = 1230.0 \, J$.

COE Prob 1c Reasoning: $KE = 0.5(5.0 \, kg)(10.0 \, m/s)^2 = 250.0 \, J$. $PE = (5.0 \, kg)(9.8 \, m/s^2)(20.0 \, m) = 980.0 \, J$. $Total = 1230.0 \, J$.

Work Prob 1 Claim: $v = 11.5 \, m/s$.

Work Prob 1 Reasoning: $F_{net} = 100.0\cos(36.87^\circ) - 0.1(mg - 100.0\sin(36.87^\circ)) = 66.4 \, N$. $W_{net} = 66.4 \, N \cdot 20.0 \, m = 1328.0 \, J$. $1328.0 \, J = 0.5(20.0 \, kg)v^2 \implies v = 11.5 \, m/s$.

Work Prob 6 Claim: $W_f = 336.0 \, J$.

Work Prob 6 Reasoning: $PE_{lost} = (10.0 \, kg)(9.8 \, m/s^2)(8.0 \, m) = 784.0 \, J$. $KE_{gained} = 0.5(10.0+4.0 \, kg)(8.0 \, m/s)^2 = 448.0 \, J$. Friction energy loss $= 784.0 \, J - 448.0 \, J = 336.0 \, J$.

AP Energy Notes

I. Energy & Work

A radiometer shows that light is energy by converting light energy into heat energy and then into movement (kinetic energy)

II. Work

II. Work Case Studies

W = Fd

D. Finding Work using a F vs. d plot

III. Potential Energy

Gravitational PE

Elastic PE

IV. Kinetic Energy & Power

I. Kinetic Energy

Power

V. Conservation of Energy

VI. Advanced COE & Work