Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance...

Physics 6A

Work and Energy examples

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Work and Energy



Energy comes in many forms. We will most often encounter two kinds of energy:

Kinetic Energy – Energy of Motion. Any moving object has kinetic energy.

The formula is KE = ½ mv2

Potential Energy – Stored Energy. There will be several types of potential energy:

* Gravitational – Energy stored by lifting an object above the earth. We will have a more

robust formula later, but for now: Ugrav = mgh

* Elastic – Energy stored by stretching or compressing a spring.

The formula is Uelastic = ½ kx2

* Electric – Energy stored by charges in an electric field. We will see this next quarter.

Work and Energy



Work is energy transferred to a moving object when a force acts on it. To do work, the force must line up with the motion of the object. Perpendicular forces do no work.

We will have two formulas involving work.

W = Fdcos(θ)

W = ΔKE

Our main concept that ties it all together is Conservation of Energy. This says that the total energy of a system does not change. We can write down a formula that accounts for all the forms of energy:

KEi + Ui + WNC = KEf + Uf

This will be the template for most of the problems you will do involving energy.



Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a force of 400N. If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on the box and the final speed when the box is pushed 10m.Assume the applied force is horizontal.



Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a force of 400N. If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on the box and the final speed when the box is pushed 10m.Assume the applied force is horizontal.

400 N

weight

Normal force

friction

Here is the free-body diagram.Since the box is moving horizontally, the only forces that do work are friction and the 400N push.

Calculate the force of kinetic friction:

𝑓 𝑘=𝜇𝑘𝑚𝑔=(0.2 ) (100𝑘𝑔 )(9.8𝑚𝑠2 )=196𝑁

Work done by each force:

𝑊 h𝑝𝑢𝑠 = (400𝑁 ) (10𝑚)=4000 𝐽𝑊 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛=− (196𝑁 ) (10𝑚 )=−1960 𝐽

Total work done on box:𝑊 𝑡𝑜𝑡𝑎𝑙=2040 𝐽

10m



Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force of 400N, applied downward at an angle of 30°.If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on the box. Does the box move?




400 N

weight

Normal force

friction30°

In the last question, we assumed the box was moving because the problem told us ho far it moved. This one is different, and we have to determine whether or not the box even moves.

To do this, we should find the maximum friction force and compare to the forward push – if the push is not enough to overcome static friction the box will not move.

First we will need to break the 400N push into components.The downward component will increase the normal force.




400 N

weight

Normal force

friction30°

First we will need to break the 400N push into components.

𝑥 : (400𝑁 ) (𝑐𝑜𝑠30 ° )=346𝑁

𝑦 : (400𝑁 ) (𝑠𝑖𝑛30 ° )=200𝑁200 N

346 N

To get the max friction we need the normal force, which will be greater because of the downward push:

𝐹 𝑛𝑜𝑟𝑚𝑎𝑙=200𝑁+ (100𝑘𝑔 )(9.8𝑚𝑠2 )=1180𝑁

𝑓 𝑠 ,𝑚𝑎𝑥=𝜇𝑠𝐹 𝑛𝑜𝑟𝑚𝑎𝑙=(0.4 ) (1180)=472𝑁

Compare this to the forward component of the push (only 346 N). Friction is strong enough to hold the box in place, so there is no motion.No work is done on the box.

Note that the actual friction force is only 346N – just enough to hold the box.



Example: A 100kg box is released from rest at the top of a frictionless 10-meter high ramp that makes an angle of 30° with the horizontal.

Find the final speed of the box when it reaches the bottom of the ramp.

Compare to the impact speed when the box is pushed over the edge and free-falls to the ground instead.

10m

30°



Example: A 100kg box is released from rest at the top of a frictionless 10-meter high ramp that makes an angle of 30° with the horizontal.


Compare to the impact speed when the box is pushed over the edge and free-falls to the ground instead.

10m

30°

We can do this one with conservation of energy.

Here is the basic format:


0 + mgh + 0 = ½ mv2 + 0

Solving for v:

𝑣=√2 h𝑔 =√2(9.8𝑚𝑠2 )(10𝑚)=14𝑚𝑠

If the box is pushed over the edge, we can use conservation of energy again, and get the exact same result. This happens because there was no friction on the ramp, so the speed at the bottom is the same in both cases.



Example: A 100kg box is released from rest at the top of a 10-meter high ramp that makes an angle of 30° with the horizontal. Assume the coefficients of friction are μk=0.2 and μs=0.3.


10m

30°





10m

30°

We can do this one with conservation of energy.

Here is the basic format:


0 + mgh + Wfriction = ½ mv2 + 0

We need to find the work done by kinetic friction as the box slides down the ramp. Also, does it even slide, or is there enough static friction to hold it in place?





30°

Does the box slide at all?We should draw the free-body diagram to see what forces are in play. The friction force is related to the Normal force.

mg

Normal forcefriction

mgcos30

mgsin30

Let’s calculate the max static friction and the downhill gravity force to see which is bigger.

𝑚𝑔𝑠𝑖𝑛30 °=(100𝑘𝑔 )(9.8𝑚𝑠2 ) (0.5 )=490𝑁

𝑓 𝑠 ,𝑚𝑎𝑥=𝜇𝑠𝑚𝑔𝑐𝑜𝑠30 °=(0.3 ) (100𝑘𝑔 )(9.8𝑚𝑠2 ) (0.866 )=255𝑁

Looks like plenty of downhill force to overcome friction. The actual friction will be kinetic:

𝑓 𝑘=𝜇𝑘𝑚𝑔𝑐𝑜𝑠30 °= (0.2 ) (100 𝑘𝑔 )(9.8𝑚𝑠2 ) (0.866 )=170𝑁

As the box slides, kinetic friction will do work against the motion.





10m

30°

Now we can fill in the conservation of energy formula.

0 + mgh + Wfriction = ½ mv2 + 0

Work done by kinetic friction:

d

𝑊 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛=− 𝑓 𝑘𝑑=− (170𝑁 ) (20𝑚 )=−3400 𝐽

𝑠𝑖𝑛30 °=10𝑚𝑑

𝑑=10𝑚𝑠𝑖𝑛30 °

=20𝑚

(100𝑘𝑔 )(9.8𝑚𝑠2 ) (10𝑚 )−3400 𝐽=12(100 𝑘𝑔)(𝑣2)

𝑣=11.3𝑚𝑠

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction.




29°

F=11N




29°

F=11N

Initially the sled is moving at 0.5 m/s, so its kinetic energy is:

J8.05.0kg4.6mvK2

sm

212

21




29°

F=11N


J8.05.0kg4.6mvK2

sm

212

21

Next we can find the work done by the boy’s pull, and add that to the kinetic energy. Remember – work always equals the change in the kinetic energy.




29°

F=11N

The force is not aligned with the motion, so we need to use the x-component to get the work.

J24.19m229cosN11W

dcosFW


J8.05.0kg4.6mvK2

sm

212

21





29°

F=11N


J24.19m229cosN11W

dcosFW


J8.05.0kg4.6mvK2

sm

212

21


Now the total KE is 20.04J. Use this to solve for the new speed:




29°

F=11N


J24.19m229cosN11W

dcosFW


J8.05.0kg4.6mvK2

sm

212

21


Now the total KE is 20.04J. Use this to solve for the new speed:

sm2

221

5.2vkg4.6

)J04.20(2v

J04.20mvK



2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis.




We will use conservation of energy for this one. Notice that the picture already has y=0 defined for us at the top of the tube (this is also the equilibrium position of the spring).

[1]

[3]





At the beginning (I will call this position [1]) the energy of the system is all potential – the ball is not moving so K=0. We need to account for the compression of the spring, as well as the gravitational potential energy of the ball:

[1]

[3]






J17.17E

m25.08.9kg5.1m25.0667E

mgykyE

1

sm2

mN

21

1

1212

11

2

[1]

[3]






J17.17E

m25.08.9kg5.1m25.0667E

mgykyE

1

sm2

mN

21

1

1212

11

2

[1]

[3]

Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the ball is not moving (K=0) and the spring is at equilibrium (Uelastic = 0).

max3 mghE






J17.17E

m25.08.9kg5.1m25.0667E

mgykyE

1

sm2

mN

21

1

1212

11

2

[1]

[3]


max3 mghE

Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J






J17.17E

m25.08.9kg5.1m25.0667E

mgykyE

1

sm2

mN

21

1

1212

11

2

[1]

[3]


max3 mghE

Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J

m17.1

8.9kg5.1

J17.17h

J17.17mgh

2sm

max

max



3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands.




We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy.

gdmghmE 21i





gdmghmE 21i

Just before the block lands, both blocks are moving so we will have kinetic energy, as well as potential energy for block 1 only (block 2 is now at y=0). Also note that both blocks move at the same speed.

2212

11f vmmghmE





gdmghmE 21i


2212

11f vmmghmE

In the absence of friction, we would just set these energies equal. We must account for friction by finding the work done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less energy than we started with, as expected (kinetic friction will always take energy away from the system).





gdmghmE 21i


2212

11f vmmghmE


dgmW

1dgmW

180cosdFW

1kfric

1kfric

fricfric

The friction work is negative because the force always opposes the motion (that is why the angle is 180 degrees).





gdmghmE 21i


2212

11f vmmghmE


dgmW

1dgmW

180cosdFW

1kfric

1kfric

fricfric

The friction work is negative because the force always opposes the motion (that is why the angle is 180 degrees).

Finally we just set our final energy equal to the initial energy, plus the (negative) friction work:

sm

2121

1k22

1k22

2121

fricif

3.1vmm

dgmgdmv

dgmgdmvmm

WEE



4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the water’s surface, the diver comes to rest. If the non-conservative work done on the diver is Wnc= -5120J, what is the depth ,d?




We will use conservation of energy. First we must define our coordinate system. Using y=0 at the lowest point achieved by the diver, we have the following expressions for the initial and final energy:

0)0(mgE

)hd(mgE

f

i





0)0(mgE

)hd(mgE

f

i

Notice that the kinetic energy is zero in both places because the speed is zero.

The final energy will be the non-conservative work plus the initial energy:

J5120)hd(mg0

WEE ncif





0)0(mgE

)hd(mgE

f

i

Notice that the kinetic energy is zero in both places because the speed is zero.

The final energy will be the non-conservative work plus the initial energy:

J5120)hd(mg0

WEE ncif

We can solve this for (d+h), then subtract out the given value for h.

m5.2m0.3m5.5d

m5.58.9kg95

J5120)hd(

J5120)hd(mg0

2sm



Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance...

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