A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest.

a) When the 3 kg weight is moving with a speed of 2.2 m/s, what is the kinetic energy of the entire system? KETOT= KEwheel+KEweight
= (1/2)(I)(w2)+(1/2)(m*v2)
=(0.5* v2)(m+1/2M)
=0.5*(2.2^2)*(3+(.5*15)) J

b) If the system started from rest, how far has the weight fallen? H = KETOT/MG
= 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m

c) What is the angular acceleration at this point?
Remember that a = αR, or α = a/R

Solve for acceleration by using vf2=vi2+2ax (vf=2.2, vi=0, x=(answer in part b)) That gives you the linear a… we want angular acceleration so we just divide our linear acceleration by the radius:

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.7 m down a θ = 33° incline. The sphere has a mass M = 4.2 kg and a radius R = 0.28 m.

a) Of the total kinetic energy of the sphere, what fraction is translational? U just divide the translational found in part (b) by KEtotal ((1/2)mv2 + (1/2)Iω2) 0.71

b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline? (1/0.7)*9.8*(3.7sin(33))*.5*4.2

c) What is the translational speed of the sphere as it reaches the bottom of the ramp? You can find the velocity by using the conservation of energy theorem thing: KEi + Ui = KEf + Uf

...Example problems involving collisions 1) On a horizontal frictionless surface a puck of mass m initially at speed u collides head-on (without rotation) with a stationary puck of mass M. Find the velocities of both puck after the collision if: i) the collision is fully elastic ii) the collision if fully inelastic. i) momentum: kineticenergy: mu = mv+MV (+ve in direction of initial u) 1 /2 m u2 = 1/2 m v2 + 1/2 M V2
2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ u v + u2 (ρ - 1) = 0 quadratic eqn: b2-4ac = 4ρ2 u2 - 4 (1 + ρ) (ρ - 1) u2 = 4ρ2 u2 - 4 (ρ2 - 1) u2 = 4u2
v = [2ρ u ± (4 u2)1/2]/{2 (1 + ρ)} = [2ρ u ± 2 u]/{2 (1 + ρ)} = u (ρ ± 1)/(1 + ρ) + ⇒ v = u , and V = 0 (no collision occurs!) - ⇒ v = u (ρ - 1)/(1 + ρ) , and V = ρ (u - v) = 2ρ2 u/(1 + ρ) e.g. ρ = 1 ⇒ v = 0, V = u . as ρ → 0 ⇒ v → - u , V → 0
ii) momentum: let ρ = (m/M) kineticenergy: ratio: then
m u = (m + M) v v = u 1/(1 + 1/ρ)
1
⇒
v = u m/(m + M)
before: K1 =
/2 m u2 ,
after:
K2 =
1
/2 (m + M) v2
K2/K1 = (1 + 1/ρ) v2/u2 = 1/(1 + 1/ρ) ⇒ v = u/2 , K2/K1 = 1/2 . as ρ → 0 ⇒ v → 0 , K2/K1 → 0
e.g. ρ = 1
2) A point mass m swings under gravity from a fixed pivot on a massless cord through an angle θ to collide with a stationary block of mass M....

...Part B
Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 \rm N over a distance of 1.00 \rm m. If her mass is 20.0 \rm kg, what distance d_2 does she slide after Zak's push ends?
Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
F= Fp-Fr
E= F*Lp= (Fp-Fr)*Lp= Fr*Lr
Lr= Lp*((Fp/Fr)-1)
Lr= 1*((125/(20*9.8*0.25))-1)= 1.6 m
Mark pushes his broken car 150 m down the block to his friend’s house. He has
to exert a 110 N horizontal force to push the car at a constant speed. How much
thermal energy is created in the tires and road during this short trip?
thermal energy is created in the tires and road
= 110 * 150
=16500 J
A 30 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long.
Using energy considerations, find the magnitude of the kinetic friction force acting on the child.
The friction force F is parallel to the slope and is constant in magnitude, so its work is
W = - F d
with d = length of the slide.
ΔU = m g Δh
Therefore:
- F d = m g Δh
F = - m g Δh / d = - 30 x 9.8 x (- 4.0) / 7.0 = 168N
A block of weight...

...KINETICENERGY
Objects have energy because of their motion; this energy is called kineticenergy. Kineticenergy of the objects having mass m and velocity v can be calculated with the formula given below;
K=1/2mv²
Kineticenergy is a scalar quantity; it does not have a direction. Unlike velocity, acceleration, force, and momentum, thekineticenergy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kineticenergy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg(m/s) 2.
Examples
1. Determine the kineticenergy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
Answer:
KE = 0.5mv2
KE = (0.5)(625 kg)(18.3 m/s)2
KE = 1.05 x105 Joules
2. If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kineticenergy?
Answer:
KE = 0.5mv2
KE = 0.5(625 kg)(36.6 m/s)2
KE = 4.19 x 105 Joules
Work-Energy Theorem
Relationship between KE and W: The word...

...
Potential Energy:
Potential energy is the stored energy of position possessed by an object.
Potential Energy Formula :
Potential Energy: PE = m x g x h
Mass:
Acceleration of Gravity:
Height:
where,
PE = Potential Energy,
m = Mass of object,
g = Acceleration of Gravity,
h = Height of object,
Examples:
1. A cat had climbed at the top of the tree. The Tree is 20 meters high and the cat weighs 6kg. How much potential energy does the cat have?
m = 6 kg, h = 20 m, g = 9.8 m/s2(Gravitational Acceleration of the earth)
Step 1: Substitute the values in the below potential energy formula:
Potential Energy: PE = m x g x h
= 6 x 9.8 x 20
Potential Energy: PE = 1176 Joules
2. On a 3m ledge, a rock is laying at the potential energy of 120 J. What will be the mass of the rock.
PE = 120 J, h = 3m, g = 9.8 m/s2(Gravitational Acceleration of the earth)
Step 1: Substitute the values in the below Velocity formula:
3. A ball of mass 2 Kg is kept on the hill of height 3Km. Calculate the potential energy possessed by it?
Solution:
Mass of the body (m) = 2kg,
Height (h) = 3 Km,
Potential Energy possessed by the body =...

...Learning Goals:
• Predict the kinetic and potential energy of objects.
• Examine how kinetic and potential energy interact with each other.
In the space provided, define the following words:
Kineticenergy-is the energy of motion. An object that has motion - whether it is vertical or horizontal motion
Potential energy-is the energy of an object or a system due to the position of the body or the arrangement of the particles of the system
Open Internet Explorer. From the FMS Jump Page. Click the Potential and KineticEnergy Skate Park link. Or type in http://phet.colorado.edu/en/simulation/energy-skate-park
Observations of KE and PE
Start your skater at the top of the track. Draw or write what happens to the skater.
|Position of Skater |Result |Possible reasons why it happened. |
|[pic] |The Skater keeps going back and fourth. |It never stops because it keeps getting KE |
| | |& PE. |
Click on [pic] and run your skater through the track again. Use this tool to help you label the spots on the ramp where there is the greatest KE and PE. Label your...

...Potential and KineticEnergy lab report
Caty Cleary
4th period
Problem statement:
How does the drop height (gravitational potential energy) of a ball affect the bounce height (kineticenergy) of the ball?
Variables:
Independent variable- drop height
Dependent variable- bounce height
Controlled variables (constants) - type of ball, measurement(unit), place bounced, and the materials used for each experiment.
Hypothesis:
If the gravitational potential energy (drop height) of the ball is increased, then the kineticenergy (bounce height) will increase because the ball will pick up speed on its way down which will cause it to apply more force to the ground, making the ball bounce higher.
Materials and Procedure:
Ball(s), meter stick, balance and a flat surface.
Procedure-
1. Tape the meter stick to the side of the table with the 0-cm end at the bottom and the 100-cm end at the top. Be sure that the meter stick is resting flat on the floor and is standing straight up.
2. Choose a ball type and record the ball type in the data table.
3. Use the triple beam balance to determine the mass of the ball and record the ball’s mass in the data table.
4. Calculate the gravitational potential energy (GPE) for the ball at each drop height. Record GPE in the data table.
5. For Trial 1, hold the ball at a height of 40...

...gravitational field except with the Principle of Equivalence.
Potential Energy
Potential energy is the same as stored energy. The "stored" energy is held within the gravitational field. When you lift a heavy object you exert energy which later will become kineticenergy when the object is dropped. A lift motor from a roller coaster exerts potential energy when lifting the train to the top of the hill. The higher the train is lifted by the motor the more potential energy is produced; thus, forming a greater amount if kineticenergy when the train is dropped. At the top of the hills the train has a huge amount of potential energy, but it has very little kineticenergy.
KineticEnergy
The word "kinetic" is derived from the Greek word meaning to move, and the word "energy" is the ability to move. Thus, "kineticenergy" is the energy of motion --it's ability to do work. The faster the body moves the more kineticenergy is produced. The greater the mass and speed of an object the more kineticenergy there will be. As the train accelerates down the hill the potential energy is converted into...

...www.studyguide.pk www.studyguide.pk www.studyguide.
1
A sphere has volume V and is made of metal of density ρ.
(a) Write down an expression for the mass m of the sphere in terms of V www.studyguide.
k www.studyguide.pk www.studyguide.pk and ρ.
......................................................................................................................................[1]
k www.studyguide.pk www.studyguide.pk www.studyguide.
(b) The sphere is immersed in a liquid. Explain the apparent loss in the weight of the sphere.
k www.studyguide.pk www.studyguide.pk www.studyguide.
..........................................................................................................................................
k www.studyguide.pk www.studyguide.pk www.studyguide.
..........................................................................................................................................
..........................................................................................................................................
k www.studyguide.pk www.studyguide.pk www.studyguide.
......................................................................................................................................[3]
k www.studyguide.pk www.studyguide.pk www.studyguide.
(c) The sphere in (b) has mass 2.0 x 10 kg. When the sphere is released, it eventually
falls in the liquid with a constant speed of 6.0 cm s .
k...