Assignments 9-10

course Phy 121

‰rȾ{TӄStudent Name:

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assignment #009

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21:08:26

`q001. Note that there are 10 questions in this set.

.You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You are also probably aware that mass is often measured in kilograms.

Here we are going to develop in terms of an experiment the meaning of the Newton as a force unit. Suppose that a cart contains 25 equal masses. The cart is equal in mass to the combined total of the 25 masses, as indicated by balancing them at equal distances from a fulcrum. The cart is placed on a slight downward incline and a weight hanger is attached to the cart by a light string and suspended over a low-friction pulley at the end of the ramp. The incline is adjusted until the cart, when given a slight push in the direction of the hanging weight, is observed to move with unchanging, or constant, velocity (and therefore zero acceleration).

The masses are then moved one at a time from the cart to the hanger, so that the system can be accelerated first by the action of gravity on one of the masses, then by the action of gravity onto of the masses, etc.. The time required for the system to accelerate from rest through a chosen displacement is observed with one, two, three, four, five, ... ten of the masses.

The acceleration of the system is then determined from these data, and the acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley. It is noted that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on the incline and the entire weight will fall freely under the acceleration of gravity.

Suppose the data points obtained for the 5 of the first 10 trials were (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2). Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and y-intercept of the line. What is your slope and what is the y intercept? What is the equation of the line?

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RESPONSE -->

slope = 925 units?

y- int = 12.8 units?

acc = 925(prop) +12.8

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21:09:03

Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points should might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points.

The slope of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2.

If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 151 cm/s^2 / (.16) = 950 cm/s^2. Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points.

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RESPONSE -->

I used a TI-92 Voyage 200 lin reg

ok

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21:09:56

`q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?

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RESPONSE -->

residual plot appears randomly scattered

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21:11:28

The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended.

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RESPONSE -->

ok

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21:11:45

`q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?

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RESPONSE -->

no, it is not the case

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21:12:15

If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0.

Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin.

In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin.

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RESPONSE -->

ok

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21:13:12

`q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?

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RESPONSE -->

force = m*a

greater mass means greater force

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21:13:35

The gravitational forces exerted on the system are exerted on the suspended masses and on the cart and the masses remaining in it.

The supporting force exerted by the ramp counters the force of gravity on the cart and the masses remaining in it, and this part of the gravitational force therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses, and this part of the gravitational force is therefore the net force acting on the mass of the entire cart-and-mass system.

The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration.

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RESPONSE -->

ok

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21:14:05

`q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result.

In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated.

In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement.

How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?

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ok

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21:14:23

The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions.

If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system.

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RESPONSE -->

ok

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21:16:06

`q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?

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RESPONSE -->

about .1

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21:16:53

Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit.

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RESPONSE -->

oops, yes

9.8 = 1*9.8

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21:17:14

`q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?

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RESPONSE -->

9.8N

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21:17:19

Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit.

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ok

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21:17:33

`q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?

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RESPONSE -->

9.8N

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21:17:37

Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg.

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ok

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21:17:55

`q008. How much force would gravity exert on a mass of 8 kg?

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RESPONSE -->

8*9.8 = 78.4N

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21:18:03

Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg.

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ok

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21:18:12

`q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?

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RESPONSE -->

20N

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21:18:18

Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons.

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RESPONSE -->

ok

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21:18:31

`q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?

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RESPONSE -->

1200*2 = 2400N

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21:18:34

This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.

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RESPONSE -->

ok

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馱}Rnf\YԜfG}P

Student Name:

assignment #010

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08:56:37

`q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force is acting on the block?

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RESPONSE -->

20N

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08:56:47

The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons.

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RESPONSE -->

ok

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08:56:59

`q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2?

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RESPONSE -->

same 20 N

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08:57:07

This depends on what forces might be resisting the acceleration of the object. If the object is accelerating on a surface of some type, then there is a good chance that a frictional force is opposing the motion. If the object is being pulled upward against the force of gravity, then more force is required then if it is sliding along a low-friction horizontal surface. If it is being pulled downhill, the force exerted by gravity has a component in the direction of motion and perhaps even less force is required. However, in every case the net force, which is the sum of all the forces acting on the object, must be 20 Newtons. The person pulling on the object must exert exactly enough force that the net force will be 20 Newtons.

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RESPONSE -->

ok

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08:57:08

`q003. If friction exerts a force of 10 Newtons in the direction opposite the motion of a 10 kg object, then how much force must be exerted by someone pulling on it to accelerate the 10 kg object at 2 m/s^2, with the acceleration in the same direction as the motion?

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08:57:44

Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted fFrict = - 10 Newtons.

To achieve the given acceleration the net force on the object must be

net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons.

In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This can be thought of as 10 Newtons to overcome friction and another 20 Newtons to achieve the required net force.

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RESPONSE -->

I missed something

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08:58:14

`q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction?

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RESPONSE -->

Fnet - fFrict = m*a

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08:58:45

If Fnet is the net force and F the force actually exerted by the person, then

Fnet = F + fFrict.

That is, the net force is the sum of the force exerted by the person and the frictional force.

We know that Fnet is +20 Newtons and fFrict is -10 Newtons, so we have the equation

20 Newtons = F + (-10 Newtons).

Solving for F we see that F = 20 Newtons + 10 Newtons = 30 Newtons.

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RESPONSE -->

Of course:

Fnet = F - fFrict = m*a

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08:59:05

`q005. If a constant net force of 12 Newtons acts on a cart of mass 6 kg, then at what rate does the velocity of the cart change?

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RESPONSE -->

2 m/s/s

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08:59:12

The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus

a = Fnet / m

= 12 Newtons / (6 kg)

= 12 kg * m/s^2 / (6 kg)

= 2 m/s^2.

We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have

(kg / kg) * m/s^2 = m/s^2.

It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation.

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RESPONSE -->

ok

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08:59:39

`q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?

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RESPONSE -->

(50-10)/20 = 2m/s/s

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08:59:45

The object will accelerate at a rate determined by Newton's Second Law, Fnet = m * a. The acceleration will therefore be

a = Fnet / m.

The net force on the object will be the sum of the 50 Newton force in the direction of motion and the 10 Newton force opposed to the direction of motion. If we take the direction of motion as positive, then the net force is

Fnet = 50 N - 10 N = 40 N.

It follows that the acceleration is

a = Fnet / m

= 40 N / (20 kg)

= 40 kg m/s^2 / (20 kg)

= 2 m/s^2.

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RESPONSE -->

ok

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09:00:17

`q007. If a force of 50 Newtons is exerted opposite to the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?

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RESPONSE -->

(-50-10)/20 = -3m/s/s

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09:00:23

If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be

net force = -50 Newtons - 10 Newtons = -60 Newtons.

The acceleration of the object will therefore be

a = Fnet / m

= -60 Newtons / (10 kg)

= -60 kg * m/s^2 / (20 kg)

= -3 m/s^2.

The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object.

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RESPONSE -->

ok

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09:02:15

`q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest?

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RESPONSE -->

vo = 20 m/s a = -20/40 = -1/2 m/s/s vf = 0

t = -20/(-1/2) = 40 s

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09:02:24

The force on the object is in the direction opposite its motion, so if the direction of motion is taken to be positive the force is in the negative direction. We therefore write the net force as

Fnet = -20 Newtons.

The acceleration of the object is therefore a = Fnet / m = -20 Newtons / 40 kg

= -20 kg * m/s^2 / (40 kg)

= -.5 m/s^2.

We can therefore describe uniformly accelerated motion of the object as v0 = 20 m/s, vf = 0 (the object comes to rest, which means its velocity ends up at 0), a = -.5 m/s^2. We can then reason out the time required from the -20 m/s change in velocity and the -.5 m/s^2 acceleration, obtaining `dt = 40 seconds.

We can confirm this using the equation vf = v0 + a `dt: Solving for `dt we obtain

`dt = (vf - v0) / a

= (0 m/s - 20 m/s) / (-.5 m/s^2)

= -20 m/s / (-.5 m/s^2)

= 40 m/s * s^2 / m = 40 s.

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RESPONSE -->

ok

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09:03:36

`q009. If we wish to bring an object with mass 50 kg from velocity 10 m/s to velocity 40 m/s in 5 seconds, what constant net force would be required?

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RESPONSE -->

a = (40-10)/ 5 = 6 m/s/s

fnet = 50 * 6 = 300N

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09:03:40

The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2.

Thus the net force required is

Fnet = 50 kg * 6 m/s^2

= 300 kg m/s^2

= 300 Newtons.

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RESPONSE -->

ok

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09:04:35

`q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what must be the mass of the object?

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RESPONSE -->

a = (0-8)/4 = -2 m/s/s

-50/ -2 = 25 kg

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09:04:42

We know the net force and we have the information required to calculate the acceleration. We will therefore be able to find the mass using Newton's Second Law Fnet = m * a.

We first find the acceleration. The change in velocity from 8 m/s to rest is -8 m/s, and this occurs in 4 seconds. The acceleration is therefore -8 m/s / (4 s) = -2 m/s^2.

The 50 Newton net force must be in the same direction as the acceleration, so we have Fnet = -50 Newtons.

We obtain the mass by solving Newton's Second Law for m:

m = Fnet / a

= -50 N / (-2 m/s^2)

= -50 kg m/s^2 / (-2 m/s^2)

= 25 kg.

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RESPONSE -->

ok

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e擜|

assignment #009

Kќ^|ͪޕ~

Physics I

11-11-2006

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08:49:04

Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?

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RESPONSE -->

w = f*'ds

f = w/'ds

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08:49:11

** Knowing the distance `ds and the work `dW we note that `dW = F * `ds; we solve this equation and find that force is F=`dw/`ds **

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RESPONSE -->

ok

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08:49:37

If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?

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RESPONSE -->

ke = w = f*'ds

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08:50:37

**`dW + `dKE = 0 applies to the work `dW done BY the system and the change `dKE in the KE OF the system.

The given force acts ON the system so F `ds is work done ON the system. The work done BY the system against that force is `dW = -F * `ds.

When you use the energy equation, this is the work you need--the work done BY the system. **

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RESPONSE -->

ok

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08:51:13

Why is KE change equal to the product of net force and distance?

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RESPONSE -->

ok

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08:51:18

** It comes from the equation vf^2 = v0^2 + 2 a `ds.

Newton's 2d Law says that a = Fnet / m.

So vf^2 = v0^2 + 2 Fnet / m `ds.

Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2.

Defining KE as 1/2 m v^2 this is

F `ds = KEf - KE0, which is change in KE. **

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RESPONSE -->

ok

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08:53:35

When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?

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RESPONSE -->

are you assuming that the net force is constant?

I don't know.

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08:54:16

** Change in KE is equal to the work done by the net force, not by the force I exert.

When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance.

If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance.

It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve.

ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. **

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RESPONSE -->

ok

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ڽψƹ違iyh

assignment #010

Kќ^|ͪޕ~

Physics I

11-11-2006

o䦄c댟w`j

assignment #010

Kќ^|ͪޕ~

Physics I

11-11-2006

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21:30:13

Query introductory problem set 3 #'s 7-12

Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.

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RESPONSE -->

'dKE = Fnet * 'ds

'dKE = m*a * 'ds

'dKE = 1/2*m(Vf-Vo)^2

'dKE = 1/2*m(Vf-Vo)^2 = m*a * 'ds

'dKE = 1/2(Vf-Vo)^2 = a * 'ds

See: Vf^2 = V0^2 + 2*a*'ds

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21:30:25

** First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds.

Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **

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RESPONSE -->

ok

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21:38:28

General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.

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RESPONSE -->

35 mi/hr * 1.609km/1mi = 56.32 km/hr

35 mi/hr * 1,609m/1mi = 56,315 m/hr

35 mi/hr * 5280ft/1mi * 1hr/3600s = 51.33 ft/s

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21:39:05

We need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds.

We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches.

1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft,

5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km.

Thus

35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr.

We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s.

The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.

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RESPONSE -->

ok

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21:41:40

Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration

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95 km/hr * 1hr/3600s * 1m/1000km = 2.64 E-5 m/s

a = (Vf-Vo)/t = (0 - 2.64 E-5 m/s)/6.2 = - 4.23 E-6 m/s/s

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21:42:07

** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s.

So change in velocity is `dv = 0 m/s - 26.3 m/s = -26.3 m/s.

Average acceleration is aAve = `dv / `dt = -26.3 m/s / (6.2 s) = -4. m/s.

So the time to come to a stop is `dt = `ds / vAve = 50 m / (12.5 m/s) = 4 s.

Acceleration is rate of velocity change = change in velocity / change in clock time = -25 m/s / (4 s) = -4.2 m/s^2.

Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is

-4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.

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RESPONSE -->

oops.

Right. 1000 m = 1 km, not the other way around.

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21:42:12

univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.

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ok

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21:42:17

** If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that

x1(t) = 25 m/s * t - .05 m/s^2 * t^2.

At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is

x2(t) = 200 m + 15 m/s * t .

The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation

25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds).

Rearranging the equation we have

-.05 t^2 + 10 t - 200 = 0.

The quadratic formula tells us that solutions are

t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 )

Simplifying we get solutions t = 22.54 and t = 177.46.

At t = 22.54 seconds the trains will collide.

Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply.

GOOD STUDENT SOLUTION:

for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt)

-10 = -.('dt)

'dt = 100

so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision.

'ds = 15 m/s(100) + 200 m

'ds = 1700 m

'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m.

The trains collide. **

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ok

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"

Good work, just a minor glitch or two.