course phy201 07-04-2006......!!!!!!!!...................................
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RESPONSE --> work divided by distance
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11:31:34 ** 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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11:33:45 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 change = net force * distance
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11:35:38 **`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 --> I see why the work was supposed to be negative
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16:03:40 Why is KE change equal to the product of net force and distance?
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RESPONSE --> KE change equal to the product of net force and distance because there has to be a net change in position of the object in order for the KE to be changed from the original value, which would not be the case if all forces cancel out.
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16:09:32 ** 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 --> I see what you were really asking now. This is a way to define work in terms of the .5*m*v^2 equation
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16:14:55 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 --> The force you exert is only one component of the net force that must be entered in to the F*d equation
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16:15:04 ** 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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\xՓi輀U{mov Student Name: assignment #010
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16:20:49 `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 because 10*2=20
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16:20:52 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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16:21:26 `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 --> 20N because 10*2=20
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16:21:32 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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16:23:21 `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 --> 30N because (10*2)+10=30
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16:23:29 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 --> ok
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16:24:33 `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=F+fFrict
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16:24:38 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
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16:26:13 `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 --> 2m/s^2 because 12/6=2
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16:26:19 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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16:29:54 `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 --> 2m/s^2 because 50-10=40, 40/20=2
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16:29:59 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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16:34:12 `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 --> -3m/s^2 because -50-10=-60, -60/20=-3
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16:34:17 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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16:42:27 `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 --> 3.16s because 20/40=2, sqrt.(-20/-2)=3.16
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17:46:07 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 --> My 1st mistake: reversing the division in the acceleration calculation. 2nd mistake: using an incorrect form of the equation vf=vo+at. All clear now.
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17:49:36 `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 --> 300N because (40-10)/5=6, 6*50=300
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17:49:39 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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17:51:31 `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 --> 25kg because 8/4=2, 50/2=25
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17:51:38 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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