Asst 10 almost

course Phy 201

Here is my work. I fixed the problems with Video Experiment 6. I also have some questions on Video Experiment 12. I did not get my question answered about how to take the quiz. Do I just download and print out a copy and then go to the testing center at BRCC? Anyway, just leave me a reply.

Right on the first experiment. Now your results work. Good job.

The qa also looks good.

See my note and resubmit the Randomized Problem, or if necessary ask more questions.

See my notes on the last experiment and resubmit it also (or ask for more explanation).

Here is my work. I fixed the problems with Video Experiment 6. I also have some questions on Video Experiment 12. I did not get my question answered about how to take the quiz. Do I just download and print out a copy and then go to the testing center at BRCC? Anyway, just leave me a reply." "Ok I figured out what I did wrong in the timing of my data. I used the formula y below, but intead of working the whole thing out I mistakenly dropped the square root of 2 off of the 'dt. This left me with a smaller time than I actually had I have corrected the figures below. 'ds^2 = .5 ( 980 cm/s^2 ) * dt^2 Ok, first of all here is my raw data: 'ds(to paper) 'ds(horizontal) 'dt(sec) vAve(cm/s) 6 cm 9 cm 0.11 81.33 13.3 cm 13.5 cm 0.16 81.94 29.7 cm 20 cm 0.25 81.24 Do the differences in the horizontal velocities seem to be the result of random experimental uncertainties and errors in timing, or does there seemed to be a definite progression in the horizontal velocities as time increases? To explain my graph, I would say that the upward curve in the first point appears to spike up. The next two points seem to come down quite a bit and then appear to level off. So for that reason I would say that the velocity seems to reach a point and then levels off. My data for horizontal velocity seemed to throw me a bit at first, but I think I see that this is because of the distance that the ball has to travel and therefore it is not the acceleration of gravity that decreases, but it is the horizontal velocity that does due to the pull of gravity on the ball. I may be wrong by saying it, but it seems that the horizontal velocity is independent of time, but it is not independent of gravitational pull. The uncertainties in my graph could be off in horizontal distance of the balls on to the paper. Since we were given the distance from the table to the paper by the instructor I will not factor those in as a point of error. I believe that I was accurate to within .2 cm so I will do a chart reflecting this. 'ds 'ds(horizontal) vAve(-.2cm) vAve(+.2cm) 6cm 9cm 80 cm/s 83.63 cm/s 13.3cm 13.5cm 83.13 cm/s 85.63 cm/s 29.7cm 20cm 79.2 cm/s 80.8 cm/s Discuss the implications of these uncertainties. As for the uncertainties, I would have to say that the line still holds pretty much true that the horizontal velocity still levels off and seems to be independent of time. “jɍ²Ôôêï†a¿ß¡œóU¿­–¶ÊŠ…xÀwé Student Name: assignment #010

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17:00:13 `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 --> That would be Fnet = m * a or Fnet = 10 kg * 2 m/s^2 = 20 Newtons.

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17:00:31 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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17:02:08 `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 --> I believe the same is true you would say Fnet = m * a or Fnet = 10 kg * 2 m/s^2 = 20 Newtons.

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17:02:27 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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17:08:03 `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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RESPONSE --> I may be overthinking this, but I would say that you would have to have a force that would add up to 30 Newtons in the direction of motion. I get that by saying that the Force of the object would be 20 Newtons and the friction would be 10 Newtons so in order to cancel out the friction you would have to have a force that be equal to the Force of the object and the friction.

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17:08:26 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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17:10:18 `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 --> Without getting to fancy I would say you could make and equation of Fnet = (m * a) + tFrict.

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17:11:11 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 --> Ok the only thing I see that I missed was that I wrote out the F as m * a.

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17:30:15 `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 --> That could be expressed as Fnet = m * a or 12 Newtons = 6 kg * a or a = 12 Newtons / 6 kg = 2 m/s^2.

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17:30:23 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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17:35:47 `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 --> That would be Fnet = F + fFrict or 50 Newtons = (20 kg * a) + 10 Newtons 40 Newtons = 20 kg * a 40 Newtons / 20 kg = a a = 2 m/s^2

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17:35:53 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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17:45:22 `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 --> Ok well I see that there is no Force stated in the direction of the object so that would mean that it is moving in the direction opposite to that of the motion of the person. I don't however think that acceleration can be negative. So therefore this is similar to the last problem where you end up with 40 Newtons / 20 kg to give you an acceleration of 2 m/s^2.

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17:47:13 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 --> One thing I was wrong about was the negative acceleration and the other was a misread by me. I thought the friction was in the opposite of the motion and I took that to mean the motion of the person and not the object. I do see how you got your answer.

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17:50:00 `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 --> Ok I will need a little help with this one. I don't know how to get the acceleration of the object.

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17:52:09 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 I see how to get it. I just needed to look at it from a different perspective.

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17:56:29 `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 --> Well first we need to look at it this way. v0 = 10 m/s and vF needs to equal 40 m/s in 5 seconds. We can then say that the 'dv = 30 m/s and 'dt = 5 s. So therefore the acceleration is going to be 6 m/s^2. Now we can solve for Fnet = 50 kg * 6 m/s^2 = 300 Newtons.

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17:56:33 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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18:01:29 `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 --> First we will have to solve the acceleration of the object. We are given v0 = 8 m/s and we know at that at rest the vF will be 0 m/s. So therefore the 'dv = 8 m/s. The 'dt = 4 s. Now we can solve for acceleration = 8 m/s / 4 s = 2 m/s^2. Now we can solve for the mass. 50 Newtons = m * 2 m/s^2 50 Newtons / 2 m/s^2 = m m = 25 kg

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18:02:40 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 I see that I missed my negative values. That is the case due the fact that we are bringing the object to a stop.

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Random Quiz 5 Version 1 A ball starting from rest rolls 11 cm down an incline on which its acceleration is 27 cm/s2, then onto a second incline 44 cm long on which its acceleration is 9 cm/s2. How much time does it spend on each incline? I may be wrong in saying this but I don't believe you can solve this problem just with the information that is given. If you don't know the final velocity then you can't determine the average velocity or the change in velocity. This problem cannot be solved as is. Please let me know if I am wrong.

You know v0, `ds and a on the first incline so you can find everything else.

You have to set everything up again for the second incline, since now the acceleration is different. vf on the first incline is v0 on the second. You know a and `ds on that incline. So you can solve it.

Here is the data for number of paper clips vs. rise of the ramp: # of P Clips rise(above)cm rise(below)cm 15 6.2 4.2 12 5.5 3.5 9 4.7 2.7 6 4.2 2.2 3 3.4 1.4 The run for all of these is 107 cm. Now here is my data for the force vs. slope table: # of P Clips Mass(grams) Slope Force(Newtons) 15 18 0.048 0.86 12 14.4 0.041 0.60 9 10.8 0.034 0.37 6 7.2 0.029 0.21 3 3.6 0.022 0.08

Again you need to explain where every column comes from and how you obtained the results. At least one of your columns does not follow from your data.

Construct an accurate graph of force in Newtons vs. effective slope. Sketch the straight line that best fits the data. Determine whether this straight line indicates a force vs. effective slope proportionality. Recall that a proportionality is indicated when, within the limits of experimental error, the straight line that fits the data passes through the origin and the data points. I constructed my graph and I do believe that the line passes through the points if not directly it is very near. I think that if it does not exactly on my graph then you could chalk this up to error in measurement. From the mass of the cart (approximately 550 grams) determine its weight. Is the weight, within experimental error, equal to the slope of the graph? I thought that the 550 grams is the weight?

Weight is the force exerted on a mass by gravity. The object has mass 550 grams wherever it is; it only has weight if it's near another massive object (e.g., a planet).

How much force does the Earth's gravitational field exert on this object? You know how much acceleration the gravitational field gives it, so you can find the force.

Can we as a result conclude that, for small slopes at least, the force required to restrain the cart from accelerating down the incline is equal to the weight of the cart multiplied by the effective slope of the incline? I'm not sure what the weight is so I'll have to wait an explanation. Can we conclude that for small slopes, the net force accelerating a cart coasting down that slope is equal to the weight of the cart multiplied by the effective slope of the incline? Once again at a loss. How can we therefore conclude that the acceleration of a low-friction cart down a small slope is equal to the acceleration of gravity multiplied by the effective slope of the incline? An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 12 cm/s^2. .206 seconds later it passes a point 7.4 cm up the ramp from its initial position. What are its possible initial velocities, and what is a possible scenario for each? What we know is the 'ds = 7.4 cm and the 'dt = .206 that would make this: vAve = 7.4 cm / .206 s = 35.92 cm/s A possible v0 could be 33.44 cm/s or a little closer to 30 cm/s. Since we are going up a ramp though you would anticipate the velocity to decrease so I would also say that it could have been about 40 cm/s. What is the maximum distance the object travels up the ramp? Given the rate at which it is traveling I would say that you could not get farther up the ramp than 40 cm. "