10-17

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course Phy201

`q001. Report your data and the procedure used to obtain it.General College Physics Students: For each object, you observed a slope on which the object would eventually come to rest if given a slight push, and another slope on which the object would continue moving if given a slight push. So you had a range of slopes, such that for some slope in the range the object would neither speed up nor slow down when given a slight push. That is, for some slope in the range, the acceleration of the object on the incline was zero.

What was this range of slopes for each object?

Domino high 0.311

Domino low 0.189

Range = 0.122

Magnet high 0.331

Magnet low 0.289

Range = 0.042

Car high 0.013

Car low 0.007

Range = 0.006

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You've reported your results, which look plausible and are probably correct, but not your data.
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University Physics Students: Report your data and your conclusions.

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`q002. A mass of 12 kg is lifted from the floor to a height of 5 meters above the floor.

How much force does gravity exert on the mass?

F = ma

F = (12kg)(9.8 m/s^2) = 117.6N

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Does gravity do positive or negative work on the mass?

neg

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What is the work done by gravity on the mass?

W = F*ds

W = 117.6N * 5meters = 588J

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The work is negative, so the correct answer is -588 Joules.
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What is the change in the gravitational potential energy of the mass?

dW = dKE

dPE = -dKE

dPE = -588J

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`dPE is equal and opposite to the work done by the conservative force.

The work done by the conservative force is -588 J.

So `dPE = +588 J.

You don't know whether nonconservative forces are acting, so you can't assert that `dPE = -`dKE. That is the case only when the total nonconservative force is zero.
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What definitions were used in solving this problem?

Work energy theorem

F = ma

W = F*ds

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`q003. The 12 kg mass in the preceding is raised from the floor until its gravitational potential energy is 2000 Joules. How high was it raised?

2000J = 117.6N * ds

ds = 17 m

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If the object is released from rest and allowed to fall back to the floor, what will be its kinetic energy when it reaches the floor, assuming that no energy is lost to air resistance?

-2000J

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In this case, how fast will it be traveling when it reaches the floor?

KE = 1/2mv^2

-2000 = 1/2 (12kg)(v)^2

Sqrt333 = sqrtv^2

V = -18.26m/s

Would the negative matter or is that because it is falling to the ground???

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Goo question.

You were asked how fast. That answer to 'how fast?' is always either positive or zero, never negative.

Had the question asked for velocity, which is a vector quantity with both magnitude and direction, appropriate answers would include any of the three answers below:

18.26 m/s downward

-18.26 m/s where upward is positive

18.26 m/s where downward is positive.
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What will be its kinetic energy when it reaches the floor if during its fall the average force of air resistance is 10 Newtons?

dKE = Fnoncon+Fcon*ds

dKE = 10N+ 117.6N(17m)

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You can't add force to work.

With respect to units you can't add a quantity whose units are N to another quantity whose units are N * m.
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dKE = 2009.2J

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In this case, how fast will it be traveling when it reaches the floor?

KE = 1/2mv^2

2009.2J = ½(12kg)(v^2)

Sqrt1004.6 = sqrtv^2

31.6m/s = v

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Had you figured out the work done by air resistance (being careful about whether the work was positive or negative) you would have gotten this.
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What definitions were used in solving this problem?

KE = 1/2mv^2

Work energy theorem

PE = -KE

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`q004. An incline rises from left to right, making an angle of 20 degrees with horizontal. A block of mass 12 kg rests on the incline.

What is the angle of the weight vector of that block relative to an x axis which is parallel to the incline?

20 degrees

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What is the weight of the mass?

12kg

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That's the mass.

The weight of the mass is the force exerted on it by gravity.
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What are the components of the weight in the x and the y directions?

mgx = cos(250) = -0.34

mgy = sin(250) = -0.94

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You have to multiply the sine and cosine of the angle by the weight in order to get the components of the weight.
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If there is no friction present, what acceleration will result from the x component of the weight?

9.8m/s^2(-.34) = 33.32

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If you have calculated the x component of the weight, then to find the corresponding acceleration you would divide the x component by the mass (a = F / m).

The result would indeed be 9.8 m/s^2 * cos(250 deg), but you need to go through Newton's Second Law to get this result.
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Why doesn't the y component of the weight result in an acceleration in the y direction?

9.8m/s^2(-.94) = -9.21

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If friction exerts a force opposing motion down the incline equal to 25% of the magnitude of the y component of the weight, then what will be the net force accelerating the block down the incline, and what will be its acceleration?

F = ma

0.235 = 12kg(a)

a = 0.02m/s^2

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If .235 was the y component of the weight this would be correct; however .235 hasn't appeared previously in this problem and it's not clear how you got it or what its units are.

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there is no acceleration in the y direction

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What definitions, laws and principles were used in solving this problem?

Mass energy theorem

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There is no mass energy theorem.

There is a work energy theorem, but it doesn't appear to have been used here, nor is it necessary to use it (though it could be applied).

You have used the definitions of the sine and cosine, the components of a vector, and Newton's Second Law.
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`q005. If the block in the preceding travels 15 meters along the incline, by how much does its altitude change?

daltitude =alt2-alt1

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A 15 meter displacement up the incline can be represented by a vector of magnitude 15 meters and angle 20 degrees relative to horizontal. Letting the x axis for the moment be horizontal, you would find that the 'rise' or y component of this vector is 15 meters * sin(20 degrees), or about 5 meters (very roughly).

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So a 15 meter displacement up the incline implies an altitude change of about 5 meters.

At the object slides 15 meters down the incline the altitude change is -5 meters.
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For the rest of the problem you want to have the x axis parallel to the incline.

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If the block slides 15 meters down the incline, is the gravitational force in the direction of motion, or only partly in the direction of motion?

Partly in the direction of motion

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What is the component of gravity in the direction of motion?

Fgrav = ma

F = (12kg)(9.8m/s^2)

F = 117.6N

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That is the entire force exerted by gravity on the mass, not the component of this force in the direction of motion.

This weight vector makes an angle of 250 degrees with an x axis directed up the incline. You need to find the x and y components of the weight vector, then use the appropriate component to find the force in the direction of motion.
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How much work does gravity therefore do on the block, and what is its change in gravitational potential energy?

W = F*ds

W = 117.6N(15m)

W = 1764 J

KE = 1764J

dPE = -1764J

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The block does not move 15 meters in the vertical direction, but if it did this result would be correct.

Per previous notes, in moving 15 meters down the incline the mass moves about 5 meters in the downward vertical direction.
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If no nonconservative forces act, what must be the change in its kinetic energy?

dKE = 1764J

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How fast will it be moving at its final kinetic energy?

KE = 1/2mv^2

1764J = ½(12kg)(v^2)

Sqrt 294 = sqrt v^2

V = 17.1 m/s

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You method on the last two questions is valid. Your energy calculation isn't correct, per previous notes.
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If you multiply the weight of the block by the altitude change you calculated in the first of the questions in this problem, what is your result?

daltitude (12kg)

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What definitions, laws and principles were used in solving this problem?

KE = 1/2mv^2

F = ma

Work energy theorem

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`q006. If the frictional force on the block in the preceding questions is equal and opposite to the x component of the weight, then what is the frictional force as a percent of the y component of the weight?

Ffriction = -(-.34)

Ffriction = .34

About 36%

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You need to base these answers on the correct x and y components of the weight.
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What is the frictional force as a percent of the normal force?

Fnorm = -(mgy)

Fnorm = -(-.94)

Fnorm = .94

About 95%

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The preceding note applies here as well.
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How will the block behave under these conditions if given a slight push down the incline?

Slide down the incline due to the gravity and the altitude of the incline where the block is resting.

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What definitions, laws and principles were used in solving this problem?

Fnorm = -(mgy)

Ffriction = -(mgx)

KE = 1/2mv^2

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`q007. Give a brief synopsis of what you learned by using the assigned PHeT program.

Mostly how to measure, add and compare vectors. There were many different exercises that were helpful. The program seemed very easy to follow when compared to others.

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Be sure to see notes here as well as on earlier assignments about multiplying the sines and cosines by the magnitudes of the vectors to get their components.

I've inserted a number of other notes also.

Let me know if you have questions.
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10-17

@& You've reported your results, which look plausible and are probably correct, but not your data. *@

@& The work is negative, so the correct answer is -588 Joules. *@

@& You can't add force to work. With respect to units you can't add a quantity whose units are N to another quantity whose units are N * m. *@

@& Had you figured out the work done by air resistance (being careful about whether the work was positive or negative) you would have gotten this. *@

@& That's the mass. The weight of the mass is the force exerted on it by gravity. *@

@& You have to multiply the sine and cosine of the angle by the weight in order to get the components of the weight. *@

@& If you have calculated the x component of the weight, then to find the corresponding acceleration you would divide the x component by the mass (a = F / m). The result would indeed be 9.8 m/s^2 * cos(250 deg), but you need to go through Newton's Second Law to get this result. *@

@& If .235 was the y component of the weight this would be correct; however .235 hasn't appeared previously in this problem and it's not clear how you got it or what its units are. *@

@& There is no mass energy theorem. There is a work energy theorem, but it doesn't appear to have been used here, nor is it necessary to use it (though it could be applied). You have used the definitions of the sine and cosine, the components of a vector, and Newton's Second Law. *@

@& So a 15 meter displacement up the incline implies an altitude change of about 5 meters. At the object slides 15 meters down the incline the altitude change is -5 meters. *@

@& The block does not move 15 meters in the vertical direction, but if it did this result would be correct. Per previous notes, in moving 15 meters down the incline the mass moves about 5 meters in the downward vertical direction. *@

@& You method on the last two questions is valid. Your energy calculation isn't correct, per previous notes. *@

@& You need to base these answers on the correct x and y components of the weight. *@

@& The preceding note applies here as well. *@

@& Be sure to see notes here as well as on earlier assignments about multiplying the sines and cosines by the magnitudes of the vectors to get their components. I've inserted a number of other notes also. Let me know if you have questions. *@

@&
You've reported your results, which look plausible and are probably correct, but not your data.
*@

@&
The work is negative, so the correct answer is -588 Joules.
*@

@&
You can't add force to work.

With respect to units you can't add a quantity whose units are N to another quantity whose units are N * m.
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@&
Had you figured out the work done by air resistance (being careful about whether the work was positive or negative) you would have gotten this.
*@

@&
That's the mass.

The weight of the mass is the force exerted on it by gravity.
*@

@&
You have to multiply the sine and cosine of the angle by the weight in order to get the components of the weight.
*@

@&
If you have calculated the x component of the weight, then to find the corresponding acceleration you would divide the x component by the mass (a = F / m).

The result would indeed be 9.8 m/s^2 * cos(250 deg), but you need to go through Newton's Second Law to get this result.
*@

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If .235 was the y component of the weight this would be correct; however .235 hasn't appeared previously in this problem and it's not clear how you got it or what its units are.

*@

@&
There is no mass energy theorem.

There is a work energy theorem, but it doesn't appear to have been used here, nor is it necessary to use it (though it could be applied).

You have used the definitions of the sine and cosine, the components of a vector, and Newton's Second Law.
*@

@&
So a 15 meter displacement up the incline implies an altitude change of about 5 meters.

At the object slides 15 meters down the incline the altitude change is -5 meters.
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The block does not move 15 meters in the vertical direction, but if it did this result would be correct.

Per previous notes, in moving 15 meters down the incline the mass moves about 5 meters in the downward vertical direction.
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You method on the last two questions is valid. Your energy calculation isn't correct, per previous notes.
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You need to base these answers on the correct x and y components of the weight.
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@&
The preceding note applies here as well.
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@&
Be sure to see notes here as well as on earlier assignments about multiplying the sines and cosines by the magnitudes of the vectors to get their components.

I've inserted a number of other notes also.

Let me know if you have questions.
*@