course Phy201 ????a|??+???g?assignment #010010. `query 10
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21:35:11 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 --> through the graph and the equation confidence assessment:
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21:35:39 ** 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 self critique assessment:
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21:40:38 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 = 56.32 km/hr, 15.65 m/s, 51.45f/s confidence assessment:
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21:40:57 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 self critique assessment:
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21:43:01 Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration
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RESPONSE --> 5.16m/s confidence assessment:
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21:43:29 ** 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 --> ok self critique assessment:
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21:43:40 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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RESPONSE --> ok confidence assessment:
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21:43:47 ** 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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RESPONSE --> ok self critique assessment:
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?????????? assignment #010 010. Note that there are 10 questions in this set. .Force and Acceleration. Physics I 09-28-2008
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21:25:41 `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 --> 20 Newtons confidence assessment:
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21:25:49 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 self critique assessment:
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21:26:11 `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 --> 20 Newtons confidence assessment:
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21:26:21 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 self critique assessment:
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21:27:06 `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 --> 30 Newtons confidence assessment:
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21:27:22 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 self critique assessment:
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21:28:12 `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 Frict = F exerted confidence assessment:
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21:28:39 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 self critique assessment:
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21:29:15 `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 --> 2m/s confidence assessment:
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21:29:21 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 self critique assessment:
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21:29:47 `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 --> 2m/s confidence assessment:
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21:29:54 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 self critique assessment:
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21:30:21 `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 --> 2m/s^2 confidence assessment:
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21:30:29 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 self critique assessment:
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21:31:40 `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 --> 780N confidence assessment:
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21:31:53 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 self critique assessment:
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21:32:38 `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 confidence assessment:
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21:32:43 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 self critique assessment:
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21:33:46 `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 confidence assessment:
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21:33:55 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 self critique assessment:
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