The Work-Energy Relationship



ENERGY LESSON 4

The Work-Energy Relationship

Internal vs. External Forces

We will learn that there are certain types of forces, which when present and when involved in doing work on objects will change the total mechanical energy of the object.

And there are other types of forces which can never change the total mechanical energy of an object, but rather can only transform the energy of an object from potential energy to kinetic energy (or vice versa).

The two categories of forces are referred to as internal forces and external forces.

|Internal Forces |External Forces |

|Fgrav |Fapp |

|Fspring |Ffrict |

| |Fair |

| |Ftens |

| |Fnorm |

 

EXTERNAL FORCE

When net work is done upon an object by an external force, the total mechanical energy (KE + PE) of that object is changed.

If the work is positive work, then the object will gain energy.

If the work is negative work, then the object will lose energy.

The gain or loss in energy can be in the form of potential energy, kinetic energy, or both.

Under such circumstances, the work which is done will be equal to the change in mechanical energy of the object.

INTERNAL FORCE

When the only type of force doing net work upon an object is an internal force (for example, gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant.

In such cases, the object's energy changes form. For example, as an object is "forced" from a high elevation to a lower elevation by gravity, some of the potential energy of that object is transformed into kinetic energy.

Yet, the sum of the kinetic and potential energies remain constant.

When the only forces doing work are internal forces, energy changes forms - from kinetic to potential (or vice versa); yet the total amount of mechanical is conserved. This is called energy conservation.

 EXAMPLES: In the following descriptions, the only forces doing work upon the objects are internal forces - gravitational and spring forces. Thus, energy is transformed from KE to PE (or vice versa) while the total amount of mechanical energy is conserved.

Read each description and indicate whether energy is transformed from KE to PE or from PE to KE.

| |Description of Motion |KE to PE or PE to KE? |

| | |Explain. |

|1. |A ball falls from a height of 2 meters in the absence|PE to KE |

|  |of air resistance. |The ball is losing height (falling) and gaining speed. Thus, the |

| |[pic] |internal or conservative force (gravity) transforms the energy from PE|

| | |(height) to KE (speed). |

| | |  |

|2. |A skier glides from location A to location B across a|PE to KE |

| |friction free ice. |The skier is losing height (the final location is lower than the |

| |[pic] |starting location) and gaining speed (the skier is faster at B than at|

| | |A). Thus, the internal force or conservative (gravity) transforms the |

| | |energy from PE (height) to KE (speed). |

| | |  |

|3. |A baseball is traveling upward towards a man in the |KE to PE |

| |bleachers. |The ball is gaining height (rising) and losing speed (slowing down). |

| |[pic] |Thus, the internal or conservative force (gravity) transforms the |

| | |energy from KE (speed) to PE (height). |

| | |  |

|4. |A bungee cord begins to exert an upward force upon a |KE to PE |

| |falling bungee jumper. |The jumper is losing speed (slowing down) and the bunjee cord is |

| |[pic] |stretching. Thus, the internal or conservative force (spring) |

| | |transforms the energy from KE (speed) to PE (a stretched "spring"). |

| | |One might also argue that the gravitational PE is decreasing due to |

| | |the loss of height. |

| | |  |

|5. |The spring of a dart gun exerts a force on a dart as |PE to KE |

| |it is launched from an initial rest position. |The spring changes from a compressed state to a relaxed state and the |

| |[pic] |dart starts moving. Thus, the internal or conservative force (spring) |

| | |transforms the energy from PE (a compressed spring) to KE (speed). |

| | |  |

When work is done by external forces, the total mechanical energy of the object is altered.

The work that is done can be positive work or negative work depending on whether the force doing the work is directed opposite the object's motion or in the same direction as the object's motion.

If the force and the displacement are in the same direction, then positive work is done on 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.

The following descriptions involve external forces (friction, applied, normal, air resistance and tension forces) doing work upon an object.

Read the description and indicate whether the object gained energy (positive work) or lost energy (negative work). Then, indicate whether the gain or loss of energy resulted in a change in the object's kinetic energy, potential energy, or both.

|Description |+ or - Work? |Change PE or |

| | |KE or Both? |

|Megan drops the ball and hits an awesome |The work is +. |Kinetic energy. |

|forehand. The racket is moving |The force is to the right and the|The applied force of the racket causes the ball to |

|horizontally as the strings apply a |displacement is to the right. F |gain speed. Thus, the external or nonconservative |

|horizontal force while in contact with the|and d are in the same direction. |force alters the kinetic energy of the ball. |

|ball. |Thus, positive work is done. |  |

|[pic] |  | |

| |  | |

|A tee ball player hits a long ball off the|The work is +. |Both. |

|tee. During the contact time between ball |The force is up and to the right |The applied force of the bat causes the ball to gain|

|and bat, the bat is moving at a 10 degree |and the displacement is up and to|both height and speed. Since the force has an up |

|angle to the horizontal. |the right. >F and d are in the |component, it contributes to a height change. Thus, |

|[pic] |same direction. Thus, positive |the external or nonconservative force alters the |

| |work is done. |both the kinetic energy and the potential energy of |

| |  |the ball. |

| | |  |

|Rusty Nales pounds a nail into a block of |The work is +. |Kinetic energy. |

|wood. The hammer head is moving |The force is to the left and the |The applied force of the hammer causes the nail to |

|horizontally when it applies force to the |displacement is to the left. F |gain speed. Thus, the external or nonconservative |

|nail. |and d are in the same direction. |force alters the kinetic energy of the nail. |

|[pic] |Thus, positive work is done. |  |

| |  | |

|The frictional force between highway and |The work is - |Kinetic energy |

|tires pushes backwards on the tires of a |The force is to the left and the |The friction force on the car causes the car to lose|

|skidding car. |displacement is to the right. F |speed. Thus, the external or nonconservative force |

|[pic] |and d are in the opposite |alters the kinetic energy of the car. |

| |direction. Thus, negative work is|  |

| |done. | |

| |  | |

|A diver experiences a horizontal reaction |The work is +. |Kinetic energy. |

|force exerted by the blocks upon her feet |The force is to the right and the|The applied force of the starting blocks causes the |

|at start of the race. |displacement is to the right. F |diver to gain speed. Thus, the external force alters|

|[pic] |and d are in the same direction. |the kinetic energy of the diver. (NOTE: there is |

| |Thus, positive work is done. |another force - gravity - which changes the diver's |

| |  |height; but the blocks are not responsible for this |

| | |height change.) |

| | |  |

|A weightlifter applies a force to lift a |The work is + |Potential energy |

|barbell above his head at constant speed. |The force is up and the |The applied force of the causes the barbell to gain |

|[pic] |displacement is up. F and d are |height. Thus, the external or nonconservative force |

| |in the same direction. Thus, |alters the potential energy of the barbell. |

| |positive work is done. |  |

| |  | |

Note that in the five situations described above, a horizontal force can never change the potential energy of an object.

Potential energy changes are the result of height changes and only a force with a vertical component can cause a height change.

The Work-Energy Relationship

If only internal forces are doing work (no work done by external forces), there is no change in total mechanical energy; the total mechanical energy is said to be conserved.

The quantitative relationship between work and mechanical energy is expressed by the following equation:

TMEi + Wext = TMEf

NOTE: the mechanical energy can be either potential energy (in which case it could be due to springs or gravity) or kinetic energy.

KEi + PEi + Wext = KEf + PEf

NOTE: the work done by external forces can be a positive or a negative work term.

To begin our investigation of the work-energy relationship, we will investigate situations involving work being done by external forces.

EXAMPLE: Consider a weightlifter who applies an upwards force (say 1000 N) to a barbell to displace it upwards a given distance (say 0.25 meters) at a constant speed. If the barbell begins with 1500 Joules of energy, what is the final mechanical energy of the barbell?

[pic]

EXAMPLE: A baseball catcher who applies a rightward force (say 6000 N) to a leftward moving baseball to bring it from a high speed to a rest position over a given distance (say 0.10 meters). If the ball begins with 605 Joules of energy, what is the final mechanical energy of the ball?

[pic]

EXAMPLE: A shopping cart full of groceries is sitting at the top of a 2.0-m hill. The cart begins to roll until it hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart and hits a parked car with an average force of 500 N. How deep a dent is made in the car (i.e., over what distance does the 500 N force act upon the can of peaches before bringing it to a stop)?

 

[pic]

Answer: The question pertains to the can of peaches; so focus on the can (not the cart).

Initially:

PE = 0.25 kg * 9.8 m/s/s * 2 m = 4.9 J

KE = 0 J (the peach can is at rest)

  The work done is (500 N) • (d) • cos 180 = - 500*d

Finally:

PE = 0 J (the can's height is zero)

KE = 0 J (the peach can is at rest)

TMEi + Wext = TMEf

0 = PEi + Wext

0 = 0.25 kg * 9.8 m/s/s * 2 m + -500*d

d = 9.8 x 10-3 m

Go through ANIMATIONS – see directions for computer and overhead

Animations found at

ANIMATION “Which path requires the most energy”

ANIMATION “HOW FAR WILL IT SKID”

Regarding the equation discussed in the ANIMATION “HOW FAR WILL IT SKID”, Stopping distance is dependent upon the square of the velocity, in the case of a horizontal force bringing an object to a stop over some horizontal distance

TMEi + Wext = TMEf

KEi + Wext = 0 J

0.5•m•vi2 + F•d•cos(Theta) = 0 J

0.5•m•vi2 = F•d

vi2 [pic]d

This means that a twofold increase in velocity would result in a fourfold (two squared) increase in stopping distance.

A fourfold increase in velocity would result in a sixteen-fold (four squared) increase in stopping distance.

 ENERGY LESSON 4 HOMEWORK

1. A car which is skidding from a high speed to a lower speed. The force of friction between the tires and the road exerts a leftward force (say 8000 N) on the rightward moving car over a given distance (say 30 m). What is the final mechanical energy of the car, if the car begins with 320 000 Joules of energy?

2. A cart being pulled up an inclined plane at constant speed by a student during a Physics lab. The applied force on the cart (say 18 N) is directed parallel to the incline to cause the cart to be displaced parallel to the incline for a given displacement (say 0.7 m). What is the final mechanical energy of the cart?

3. Test your understanding by predicting the stopping distance values in the table below for a car with a horizontal force acting on it to bring it to a stop.

|Velocity (m/s) |Stopping Distance (m) |

|0 m/s |0 |

|  | |

|5 m/s |4 m |

|  | |

|10 m/s |  |

|  | |

|15 m/s | |

|  | |

|20 m/s | |

|  | |

|25 m/s | |

|  | |

4. A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car.

[pic]

 5. At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount.

[pic]

HOMEWORK KEY

1. –80000 J

2. 12.6 J

3. 16 m, 36 m, 64 m, 100 m

4. 39.1 m

5. 56 250 N

ENERGY LESSON 4 HOMEWORK

1. A car which is skidding from a high speed to a lower speed. The force of friction between the tires and the road exerts a leftward force (say 8000 N) on the rightward moving car over a given distance (say 30 m). What is the final mechanical energy of the car, if the car begins with 320 000 Joules of energy? [pic]

2. A cart being pulled up an inclined plane at constant speed by a student during a Physics lab. The applied force on the cart (say 18 N) is directed parallel to the incline to cause the cart to be displaced parallel to the incline for a given displacement (say 0.7 m). What is the final mechanical energy of the cart?

[pic]

3. Test your understanding by predicting the stopping distance values in the table below for a car with a horizontal force acting on it to bring it to a stop.

|Velocity (m/s) |Stopping Distance (m) |

|0 m/s |0 |

|  | |

|5 m/s |4 m |

|  | |

|10 m/s |  |

|  | |

|15 m/s | |

|  | |

|20 m/s | |

|  | |

|25 m/s | |

|  | |

Answer: (a) d = 16 m

From 5 m/s to 10 m/s is a two-fold increase (10 / 5 = 2) in velocity. Thus there should be a four-fold increase in stopping distance. Multiply 4 m by 4.

(b) d = 36 m

From 5 m/s to 115 m/s is a three-fold increase (15 / 5 = 3) in velocity. Thus there should be a nine-fold increase in stopping distance. Multiply 4 m by 9.

 (c) d = 64 m

From 5 m/s to 20 m/s is a four-fold increase (20 / 5 = 4) in velocity. Thus there should be a 16-fold increase in stopping distance. Multiply 4 m by 16.

(d) d = 100 m

From 5 m/s to 25 m/s is a five-fold increase (25 / 5 = 5) in velocity. Thus, there should be a 25-fold increase in stopping distance. Multiply 4 m by 25.

4. A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car.

[pic]

 

Answer: Initially:

PE = 0 J (the car's height is zero)

KE = 0.5*1000*(25)^2 = 312 5000 J

Finally:

PE = 0 J (the car's height is zero)

KE = 0 J (the car's speed is zero)

 

The work done is (8000 N) • (d) • cos 180 = - 8000*d

Using the equation,

TMEi + Wext = TMEf

312 500 J + (-8000 • d) = 0 J

Using some algebra it can be shown that d=39.1 m

 

5. At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount.

[pic]

Answer:

Initially:

PE = 0 J (the car's height is zero)

KE = 0.5*6000*(20)^2 = 1 200 000 J

 

Finally:

PE = 0 J (the car's height is zero)

KE = 0.5*6000*(5)^2 = 75 000 J

 

The work done is F • 20 • cos 180 = -20•F

Using the equation,

TMEi + Wext = TMEf

1 200 000 J + (-20*F) = 75 000 J

Using some algebra, it can be shown that 20*F = 1 125 000 and so F = 56 250 N

 

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download