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 w = 15.0 N sits on a frictionless inclined plane, which makes an angle θ = 23.0° with respect to the horizontal, as shown in the figure. A force of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

Part A
The block moves up an incline with constant speed. What is the total work WTotal done on the block by all forces as the block moves a distance L = 3.40 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest....

...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...

...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 .
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(i) For this sphere travelling at constant speed, calculate
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1. its kineticenergy,
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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.
kineticenergy = ...................................... J
k www.studyguide.pk www.studyguide.pk www.studyguide.
2. its rate of loss of gravitational potential energy.
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.
k www.studyguide.pk www.studyguide.pk www.studyguide.
rate = ...................................... J s
[5]
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(ii) Suggest why it is possible for the sphere to have constant kineticenergy whilst
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losing potential energy at a steady rate.
...................................................................................................................................
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...relate force and acceleration, which are key concepts in roller coaster physics. At amusement parks, Newton's laws can be applied to every ride. These rides range from 'The Swings' to The 'Hammer'. Newton was also one of the developers of calculus which is essential to analyzing falling bodies constrained on more complex paths than inclined planes. A roller coaster rider is in an 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, "kinetic...

...WHAT IS TORQUE?
Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'. The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by 'r'. Note that this distance, 'r', is also a vector, and points from the axis of rotation to the point where theforce acts. (Refer to Figure 1 for a pictoral representation of these definitions.) |
Figure 1 Definitions |
Torque is defined as
= r x F = r F sin().
In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F.
Using the right hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector.
Imagine pushing a door to open it. The force of your push (F) causes the door to rotate about its hinges (the pivot point, O). How hard you need to push depends on the distance you are from the hinges (r) (and several other things, but let's ignore them now). The closer you are to the hinges (i.e. the smaller r is), the harder it is to push. This is what happens...

...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. Assuming a fully elastic collision find the distance the block will slide if the dynamic friction is µ. during swing down m acquires speed u: m g Δh = 1/2 m u2 ⇒ u = [2g R(1 - cosθ)]1/2 let ρ = (m/M) then speed of block just after impact is: V = 2ρ2 u/(1 + ρ) v = u (ρ - 1)/(1 + ρ)
so after collision, block M has KE:
K = 1/2 M V2...

...
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 = m × g × h
Where g = 9.8 m/s2
∴ Potential Energy = 2...

...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 cm, drop the...

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