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Work and Energy Flashcards

1
Q

Define work and energy. What is the relationship between the two?

A

Work is a transfers of energy while energy if the ability to do work. Both are dependent on one another in definition and can not be defined without another

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2
Q

Mathematically define mechanical work.

A

Work = Force(distance)cosθ OR F*d

This is the work done by the force done. Therefore W is the amount of energy that the force, F, is giving to the object.
F refers to the size of the particular force doing the work.
D is the displacement of the object, how far it is movED while the force exerted on it
Cosθ = this is the angle at which the force is exerted to the object

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3
Q

Camilla runs against a block with a force of 10N and moves the block 7 meters. What is the work she exerted?

A

W = Fd = 10N7m = 70 N*m or 70 J

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4
Q

True or false: Joules is equal to N/m

A

False. Joules is the equivalent to Nm. Joules is the unit used to describe energy while Nm are units used to describe Work

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5
Q

For mechanical work, what is the mathematical energy involved with a mass moving at a velocity

A

This is kinetic energy!! KE is the energy due to motion or WORK needed to accelerate a stationary object from rest to its current velocity. KE = ½ mv^2

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6
Q

What is one necessary condition in order to define kinetic energy in terms of mechanical work?

A

The object must be at rest initially!! Therefore the kinetic energy is the energy due to the motion of work needed to accelerate a stationary object from rest to its current velocity

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7
Q

What is the energy exerted by a work of force moving a 5 kg mass at 7 m/s?

A

KE: ½ mv^2 = ½ (5 kg)(7m/s) = 122.5Nm or 122.5J

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8
Q

What is work and energy not dependent on?

A

These two physical quantities do not account for dependent variables such as: the pathway of the scenario, doesn’t account for friction, or the time it took.

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9
Q

In considering a scenario in which an object is either pushed up into the sky or is falling from the sky, what is considered as the natural force in play here?

A

Gravitational motion.

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10
Q

Assume an elevator is moving up a building at a constant velocity. What are the forces and work at play here?

A

A constant velocity means there is not overall net force therefore all the work done by the mechanism pulling on the elevator up and the work done by gravity are equal to one another.

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11
Q

Assume an elevator is accelerating up a building. What works are at play here? What is the network?

A

The work of the elevator pull and the work due to gravity are playing a role in this scenario. The work of the mechanism is greater than the work done by gravity, leading to an acceleration of the elevator

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12
Q

Find the work needed to move a 10 kg elevator 100 meters up a building.

A

Force of gravity 98N
Distance 100 m
Work = 9800 J need to move the elevator up 100 m

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13
Q

Understanding the law of conservation, where does the kinetic energy of an elevator moving up a building go?

A

Based on the Law of Conservation of Energy - Energy cannot be created nor destroyed. It gets transferred from one form to another - the kinetic energy of this mass is converted into Gravitational PE - mgh

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14
Q 
If a 9.76kg ball is raised to a height of 10.34 m, what is the potential energy of the ball in joules? 
A 675.8 
B 865.3 
C 989.0
D 1143.6
A

PE = mgh
= 9.76 kg10.34 m9.8m/s^2
= 989.0 J

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15
Q

Compare and contrast when the work done on an object will be positive or negative.

A

+ Work means that the force done on the object is in the direction of the object, therefore there is a displacement. - work means that the the force is taking energy from the object, which can occur when the force is working against the direction of the displacement of the object.

If positive work is done on an object by an external force, then the object gains mechanical energy. If the force and the displacement are in the opposite direction, then negative work is done on the object; the object subsequently loses mechanical energy

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16
Q

What does it mean for a force to have a Cos 90 to the object’s displacement. Explain in terms of work.

A
  1. When Cos 90, therefore 0, this means that the force exerted on the object is perpendicular to the displacement of the object. Perpendicular forces, do not play a role in the work done on the object, therefore W = 0.
  2. This can also mean that the object was not displaced! Therefore causing the equation to be 0 as well.
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17
Q

Describe the net work done on an object in terms of velocity.

A

W_net = 1/2mv_f^2 - 1/2mv_i^2. This is the net change of kinetic energy!!

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18
Q

What are a few assumptions you have to make when using the Work Energy Principle?

A

When using W_net = 1/2mv_f^2 - 1/2mv_i^2 or KE_f - KE_i or ΔKE, you have to assume that acceleration is constant, as well as the force exerted on the object is constant as well.

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19
Q 
Wnet = + 
A. the object is going to speed up 
B. the object is going to slow down
C. there is no change to the object’s acceleration
D. There is no change
A

A . the object is going to speed up because the object is gaining kinetic energy over time

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20
Q 
ΔKE = - 
A. the object is going to speed up 
B. the object is going to slow down
C. there is no change to the object’s acceleration
D. There is no change
A

B. the object is going to slow down. KEnet = Wnet, therefore the KE is decreasing overtime

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21
Q 
Constant acceleration of an object … 
A. Wnet > 0 
B. ΔKE < 0
C. 1/2mv_f^2 - 1/2mv_i^2 = 0 
D. KE_f - KE_i < 1
A

C. 1/2mv_f^2 - 1/2mv_i^2 = 0 . This is the same as saying KE_f - KE_i = 0, Wnet = 0, ΔKE= 0 All mean there is no change in the kinetic energy of the object therefore it is at a constant acceleration.

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22
Q

A 4kg trash can. You decide to tie string to it and pull it at 50 N. The kinetic friction on the trashcan is 30N as the trash can moves. It moves in 10m. Find the work done by each force working in this scenario.

A

There are a few forces working here: Normal force, gravitational force, pull, and frictional force
W = Fd cosθ
Work by pull = 50N10 m cos 0 = 500Nm
Work by Friction = 30N10m cos(180) = -300 N*m
Work by gravitational force = -9.8m/s^2(4kg)(10m) cos(90) = 0
Work by normal force = +9.8m/s^2(4kg)(10m) cos(90) = 0

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23
Q

Find the speed of the a 4 kg bumper car moving if the net work is 200J

A

Use the work energy principle to solve for velocity - WEnet = ΔKE
+200 J = 1/2mv_f^2 - 1/2mv_i^2.
Assuming that the kinetic energy of the trash can starts at rest => 200J = 1/2mv_f^2
Therefore v = 10m/s

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24
Q

In a scenario in which a trash can is lifted off the ground by a garbage truck, the work of the truck on the trash can is 39.2J. If the force of gravity onto the trash can is also 39.2 J, why is the trash can able to move?

A

The net work is 0, but this does not imply there is no motion! The object can still be in motion, but not gain any net force or net speed in the process, leading to a net 0 for work.

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25
Q

T/F - When work on an object is 0, this means the object is not moving

A

False. This can be! BUT! This can also mean that the force exerted onto the object is constant and therefore the acceleration is also constant! This means the object can still be in motion, but not gain any speed in the process.

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26
Q

Friction acting on a biker is 60 N as they are moving 100 N. How does this change their final velocity?

A

Have to subtract the frictional work from the final work leading to finding a network. Then you utilize the work-energy principle to solve for the final velocity of the biker

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27
Q

Temperature changes can be experienced as an object moves. What types of energy transfer of this form exists?

A

3: Conventional, convectional, and radiation.

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28
Q

What is the internal energy of a system?

A

The summation of all the different types of energies of an indv molecule and atoms found within this box. Contains both the kinetic and potential energy of all the indv molecules therefore it focuses on the microscale of energy

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29
Q

As a box is traveling at an initial velocity on top of a table and comes to a stop at the end of the table. What has caused this box to stop assuming the temperature has not changed and given that it does not change its height?

A

Because the system is unable to conserve its energy by transferring its kinetic energy into potential energy, the KE is converted into work done by friction force

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30
Q

As a box is traveling at an initial velocity on top of a table and comes to a stop at the end of the table. What is the purpose of friction force in this scenario?

A

Frictional forces act to transform energy into internal energy of our system. Therefore the kinetic energy of our block is used to increase the internal energy of the box and the table.

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31
Q

In scenarios in which energy is not conserved into other forms of energy and is lost in forms of friction or air resistance, is the law of conservation being obeyed?

A

No, even though frictional forces such as friction and air resistance do have the ability to do work and therefore is a form of energy, when the initial energy is converted into these resistive energies, we consider them lost to the environment and therefore energy is not conserved. In order for energy to be conserved, the initial energy has to be converted into other useful energies.

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32
Q

What is Hooke’s Law dependent on?

A

Hooke’s Law states the restorative force of a spring is dependent on the the spring’s change in displacement from normal resting state. Therefore it must have an initial position and a final position

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33
Q

You have a spring attached to a wall and compress it inwards with a 5N force. It moves 10 meters inwards until it stops with the same amount of compression. What is its k constant?

A

F = -kΔx. F is the restorative force and is equal to the amount of force compressing/stretching it at the point it doesn’t move anymore. Therefore F = 5N. 5N = -kx where x = 10m => k = ½

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34
Q

What variable(s) of Hooke’s law are able to change?

A

The restorative force (F) and the displacement of the spring is allowed to change. These are dependent on the force compressing it

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35
Q

Based on Hooke’s law, what does k mean?

A

K is the spring constant which is intrinsic to the spring. This entails how stiff a spring/flexible a spring is. Therefore the higher the number the ore stiff, while the less the k, the more flexible/not stiff the spring is

36
Q

Depending on the textbook, the restorative force would either be set to a formula with a negative sign or not. What is the importance of this and why would it be left out?

A

The negative sign implies the restorative force opposes the applied force. However it is left out of equations because it is not always talked about in terms of applied force. You have to be conscious that the restorative acts against the applied force however!

37
Q

How would you graph hooke’s law?

A

On a restorative force vs distance graph, a linear curve would be present. This line represents k. This means that the force applied is proportional to the restorative force as long as the motion in which the spring moves is consistent

38
Q

Compare and contrast mechanical work and spring work

A

Mechanical work is is Fd while the spring work = kx^2/(2)

39
Q

Graphically relate the work done by a spring to Hooke’s law.

A

The area under the k curve is the amount of work achieved by the spring done over the certain amount of displacement.

40
Q

How do we express work in terms of springs?

A

If a force is applied onto a spring, this is termed as work done ON a spring. This is expressed in W = ½ (k*x^2). The spring can do work, however energy comes from the work initially done onto the spring itself. We call this elastic/spring potential energy

41
Q

Oftentimes physicists take the integral of the work done on a spring. Why would they chose to take this route when one can sufficiently find the answer through it’s discrete equation of ½ kΔx^2.

A

Because the applied force causes the spring to change position over time until it reaches its maximum restorative force for the applied force, this change causes uncertainty in discrete work values. Therefore taking the integral allows precision in the value.

42
Q

As you are playing with Angelie, you guys find a slinky in her closet. As she stretches, the slinky becomes thinner and thinner, and once she lets go of the slinky, it bounces back into shape, snapping from her hands. What allows the object to be stretched in this way?

A

Its elasticity or the ability to be stretch allows the object to stretch to some extent

43
Q

Say that the amount of stretch the slinky goes through is defined as Δx as it moves from location 0 to to its final location. How can you give its relationship to the amount of force Angelie is applying to it?

A

Hooke’s Law: F = kΔx
F = the force acting on the object
k = the proportionality constant which depends on the type of material the spring is

44
Q

You and Juan are arguing about when you can use Hooke’s Law. He claims it can be used for anything that has the ability to move from initial point to point final. Is he correct?

A

While he is correct that the object must move from one location to another to obey Hooke’s law, it also has to center around objects with elastic properties (or the ability to stretch; this is defined by k)

45
Q

Say that you attached a flimsy rope to an orchard as you plan to make a swing for you and Camilla. While the rope is hanging from the tree, what is the force acting on it?

A

F = kΔx Because there is no motion or movement of the rope, the total force acting on the rope is 0

46
Q

Upon attaching a tire to complete the swing, you place Camilla on it. This causes the tire to be more than 2 centimeters from its original swinging place. What is the spring constant of the rope if Camilla only weighs 10 kilos?

A 
Weight/Force = 10kg(9.8 m/s^2) = 98N 
F = kΔx
F/Δx = k
k = 98N/2e-3 m = 4.9 e4 kg/s^2
47
Q

Observing the graph of a spring, what should you be noting?

A

X axis - elongation (Δx)
Y axis - Force (N)
Slope of curve - k (note: the value does not change as linear slopes do not change)
Elastic region, Plastic region, and Breaking point

48
Q

Catalina finally reaches the swing set after a few seconds and jumps on top of everyone. As she lands, the ropes snap and everyone falls down. Describe what just happened on a newton vs meters graph

A

With Catalina’s weight, we exceeded the ultimate strength (The max force that can be applied before breakage). This caused the k constant to enter into the breakage point meaning the rope has snapped from all the force applied.

49
Q

Picking Camilla up and resting on her hips as Juan and Catalina get up, you assess what has happened. You think back to where the set up went wrong. Had you removed Camilla earlier, how should the set up respond? If you and Juan got up before Catalina jumped on, how would the set up respond then?

A

When Camilla was on the swing set, the set up was in the elastic region, meaning that if she, the force, was removed from the setup, then the swing set would’ve returned to its original shape and form. However because with you and Juan’s additional weights on the set, the set up entered the plastic region. Therefore, even if you two got up the setup would not recover from the force and would remain the same position that you two induced on it

50
Q

Apply Hooke’s law to a spring as it is being compressed. In its compressed state, what does the force represent in this scenario?

A

During compression, the calculated force can be one of the two depending on the direction of the force: restorative force and force of compression. The force of compression causes the spring to compress, creating the restorative force, a potential force that once released will cause the spring to restore itself back into its original state. Both are equal to the k*x

51
Q

Define the relationship between the force of compression and the restorative force of spring when a force compressing a spring reaches a point where the spring is not moving anymore?

A

Fc = Fr (but in opposite directions)

52
Q

Compare and contrast the curve of a spring being stretched and a spring being compressed. How does these two differ on a force and elongation graph

A

Both are the same graph. They have a linear curve (k) on a force vs meter graph.

53
Q

Upon closer analysis of a k constant’s curve on a force vs elongation graph, you find that the area under the curve is quite representative of…

A

The area under the curve is the work the spring could do/the work done onto the spring.

54
Q

A spring with k = 10, and delta x = 5. By the time it is compressed, how much potential energy is in this spring. What does the area under this curve represent?

A

1/2 k*x^2 = ½(10)(5)^2 = 125 J
Therefore PE of a spring is simply the area under the curve of a spring compression or stretch!!! This is the work of the spring compressed at this change of distance

55
Q

Alex and you are designing a hot wheels track and he decides that there should be a loop in the middle of this one way road. He produces a 2 m high loop from the ground while you place a spring at the start of a toy race track. When you compress this spring backwards and it hits the car, this will power the small hot wheels through the track from start to finish. Design an equation of the energies present in this set. Assume this is a conservative system.

A

Because the only force input into the system is the force of compression of a spring, the generated energy is potential energy of a spring PEi = ½ kx2 (PE of a spring)
Total energy does not change. The energy is simply converted into another form
Ignoring all the minor points of conversion this initial potential energy can be converted into and focusing on only the biggest changes - at the top of the loop - the 2 energies at play here are: KE + PEf
Therefore overall equation of the set up: PEi = KE + PEf

56
Q

Calculate the potential energy of a mass of 500kg raised 4 meters above the ground.

A

PEg = mgh
= 500kg9.8m/s24m
= 19,600J

57
Q

Nonconservative forces include

A

Friction, air resistance, turbulence

58
Q

In terms of conservation, what is the result of frictional force?

A

frictional force creates other energies like thermal energy. Increase in frictional energy, increases thermal energy

59
Q

2 lifters are benching the same amount of weight, 135 lb. However one lifts faster than the other. Are both doing the same amount of work? What physical quantity do both differentiate in?

A

Yes are doing the same amount of work as work is the transfer of energy because W = f*d cosθ. both weightlifters are giving the same amount of gravitational potential energy. Both lift to the same height and drops it to the same height as well with the same mass as well
However, both differ in terms of power - this is the rate at which someone or something does work - Work/time

60
Q

True or false: Power is a conservative force

A

False, power is a nonconservative force as it is dependent on time. Power = work/time

61
Q 
Units of Power 
A. sec/Joules
B. Tesla
C. Joules/sec
D. Joules
A 
C. Joules/sec Remember Power = Work/time = Joules/sec = Watts 
Other units: 
Btu/min: watts
Horsepower (metric): foot lbf/sec
Horsepower(electric): kgf m/sec
Btu/hour: kilowatts
62
Q

1000kg car from rest takes 2 seconds to reach 5 m/s. What is the power produced by this car?

A

P = J/s AKA Work/time

Force exhibited in this case - kinetic energy
KE = 1/2mv2
= ½ (1000kg) (5m/s)2
= 500*25
= 12,500 J
This means that 14,000J is the work achieved by this car as well!

P = 12,500J/2s
= 6250 J/s AKA 6250 W

63
Q

When does the instantaneous equal the average?

A

When the physical quantity does not change at different instances of time, such as the power of a moving car does not change. At this point, instantaneous power = average power
Average power over any time interval is going to equal the instantaneous power at any moment

64
Q

What is the instantaneous speed of a moving car if it exerted a power of 500 W and a force of 100N.

A 
P = Work/time = F*d*cosθ/t 
d/t = allows isolation of these 2 and can be rewritten in terms of speed and therefore instantaneous velocity
Therefore P = F*v*cosθ
500W = 500J/sec = 500kgm2/s2
100N = 100kgm/s
θ = 0 => cos(0) = 1  
v = (500kgm2/s2*1) / 100kgm/s  
v = 5m/s is the instantaneous velocity
65
Q 
One side of a seesaw carries a 21kg mass four meters from the fulcrum and a 25.5kg mass two meters from the fulcrum. To balance the seesaw, what mass should be placed nine meters from the fulcrum on the side opposite the first two masses?
A. 15kg
B. 36kg
C. 12kg
D. 45kg
A

A. 15kg
For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides. (note: Torque is the rotational force!)
τ=Fd=mgd
The total torque must be equal on both sides in order for the net torque to be zero.
τ1+τ2=τ3
Substitute the formula for torque into this equation.
m1gd1+m2gd2=m3gd3
Now we can use the given values to solve for the missing mass.
(21kg)g(4m)+(25.5kg)g(2m)=m3g(9m)
The acceleration form gravity cancels from each term.
(21kg)(4m)+(25.5kg)(2m)=m3(9m)
84kg⋅m+51kg⋅m=m3(9m)
135kg⋅m=m3(9m)
m3=15kg

66
Q

On a seesaw, how can you define the force seen on each side?

A

The force on each side of the seesaw can be defined by Torque, which is a rotating force. Traditionally it is defined as τ = radiusForcesinθ = rFsinθ
But in simple machine, seesaw, τ is also defined as massgravitational forcedistance from fulcrum = mgd

67
Q

On a seesaw the work input is equal work output. What is the work input known as

A

This is known as the moment = force*distance this tells us how much the object will rotate and which direction therefore a twist

68
Q

100N object. Max strength to lift this object is only 10N. Therefore want to produce a force that will lead to a 10x initial force. Therefore what is the ratio for the input/output distances

A 
F1D1 = F2D2 
F1 = 10N
F2 = 100N 
D1/D2 = F2/F1 
D1/D2 = 100/10 = 10

Therefore, the distances can be any length as they have the relationship of 10.
Such as, D1 can equal 5m, but D2 has to be 50m
OR if D2 is 100m then D1 has to be 10m

69
Q

What is the trade off in mechanical advantage?

A

Though you may not have to exert as much force, you Have to push much longer in order to move the 100N object to the desired height.

70
Q

You Want to push down on the right side of a seesaw. Dr = 35m and Dl = 5m. With pushing down of 7N, how heavy can the object on the right side be?

A

FLDL = FRDR
FL = FRDR/DL
= 7N35m / 5m
= 77 = 49N

71
Q

What has gained and given up in a pulley system?

A

A reduction of force at the cost of increased distance to achieve a given value of work or energy transference. This allows the heavy objects to be lifted using a much reduced force.

72
Q

Idealized pulley tends to have what assumptions

A

Massless and frictionless. With these conditions, the work in will = the work out. But this is not true on all machines in reality, therefore no 1000% efficiency

73
Q

How do you figure out the efficiency of a machine with mechanical advantage? Apply this to a pulley system.

A

Efficiency = Wout / Win
= foutdout/findin
= (load)(load distance)/[Effort*effort distance)]
Where the load - weight (force) is the mass on the balance; load distance - the distance the weight moves | effort - force put into the system | Effort distance - amount of rope pulled

74
Q

Apply conservation of energy to mechanical advantage.

A

The work in has to equal the work out (ideally)
Wi = WO
FIDI = FODO

75
Q

At which point can you relate the average velocity to power?

A

When the power is average power as well.
Pave = W/t
Pave = Fd/t
Pave = Fvave

76
Q

What is the equation for work in terms of force?

A

Work = Force x Distance

77
Q

If Johnny pushes a box with a force of 21.24 N over a distance of 10.76 m, how much work has Johnny done to the box?

(A) 165.98
(B) 228.54
(C) 302.67
(D) 452.12

A

(B) 228.54
Work = Force x Distance
Work = 21.24 N x 10.76 m
Work = approx. 200 (actual: 228.54)

78
Q

What is the equation for kinetic energy in terms of velocity?

A

Kinetic Energy = (1/2)mv^2
m = mass
v = velocity

79
Q

What is mechanical advantage?

A

This is a measure of the increase in force accomplished by using a tool | magnitude of the force exerted on an object by a simple machine (Fout) to the force actually applied on the simple machine MA = Fout / Fin
Examples: inclines, pulleys, seesaw, wheels, lever, screw.
MCAT focuses on plane, lever, and pulleys the most

80
Q

What are the units in describing mechanical advantage?

A

Unitless! Fout / Fin = no units

81
Q

A block weighing 100N is pushed up a frictionless incline over a distance of 20m to a height of 10m as shown below. Find the minimum force required to push the block.

A

Sinθ = S = O/H = 10/20
F = mgsinθ
= 100N(10/20) = 50N

Note: 10/20 = sinθ not sin(½)!!!
Remember that in an incline, the force of gravity acting on the object is only the y component

82
Q

A block weighing 100N is pushed up a frictionless incline over a distance of 20m to a height of 10m as shown below. Find the force required and the work done by the block were simply lifted vertically 10m.

A

To raise the block vertically, an upward force equal to the object’s weight (100N) would have to be generated. The work done by lifting force is
W = Fdcosθ
= 100N10mcos(1) = 1000J

83
Q

Contrast mechanical advantage from efficiency of machines.

A

Mechanical advantage is a property of a machine and is the factor by which it multiplies any applied force
Efficiency of any machine measures the degree to which friction and other factors reduce the actual work output of the machine from its theoretical maximum

84
Q

Recall the units used for work.

A

Work is the term used to describe the process of energy transfer. This means that work shares the same units as energy and since energy is defined as joules, work is also energy
Joules = kg*m2/s2

85
Q

Work and energy share the same units, and the definition of one is important to define another. Are they the same quantities?

A

No, while both share units of Joules, these are not the same definition

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