course Phy 231 yx||}^Kacsxְassignment #015
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20:44:55 Set 4 probs 1-7 If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?
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RESPONSE --> If we know Fnet and 'dt we can use those to find Fnet * 'dt = Impulse or change in momentum. confidence assessment: 3
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20:45:08 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
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RESPONSE --> Ok self critique assessment: 3
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20:45:42 What is the definition of the momentum of an object?
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RESPONSE --> Momentum is the product of the mass of an object and its velocity or m * v = p confidence assessment: 3
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20:46:58 ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **
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RESPONSE --> Ok self critique assessment: 3
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20:47:36 How do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?
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RESPONSE --> We find the impulse or change in momentum by the product of Fnet * 'dt = 'dp. confidence assessment: 3
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20:47:57 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
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RESPONSE --> Ok self critique assessment: 3
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20:53:38 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
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RESPONSE --> Newtons's Second law is the concept expressed by Fnet = m * a; therefore, when we have Fnet * 'dt is can also be shown as m * a * 'dt, which is equal to m * 'dv when the object begins froms rest. This is how the equation of momentum can be used as an easier way to calulate the changes in velocity when only force and time is known since for non-calculus based course a change in momentum implies a change in velocity only since mass does not change. confidence assessment: 3
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20:53:56 ** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt **
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RESPONSE --> Ok self critique assessment: 3
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20:58:31 If you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies?
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RESPONSE --> First we can use the method that using the equations of motion that we first learned in this course, which would to first use the time interval and 'dv to find acceleration with the equation (vf - v0)/ 'dt = a, and then finding the product of a and m to find m * a = Fnet. The second method involvind the impulse-momentum theorem would be to take the product of mass and 'dv to find momentum and then divide by 'dt to find (m * 'dv)/ 'dt = Fnet. The quantities, units, and general flow of information is the same, but they can be applied to different scenarios to find the desired value. confidence assessment: 3
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20:58:44 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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RESPONSE --> Ok self critique assessment: 3
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21:03:43 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
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RESPONSE --> Since Fnet * 'ds = 'dW = KE, which is the value of KE of a system or object, we can start by breaking down the values used to determine Fnet and 'ds. First we know Fnet = m * a, second we know that the equation for 'ds = a * t^2 when there is no initial velocity. So we can work backworks from this equation to what we began with and find out the value of energy. confidence assessment: 3
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21:04:30 ** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **
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RESPONSE --> Ok, I used the format for work instead the derived equation for KE involving mass and velocity. self critique assessment: 3
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21:09:25 What is kinetic energy and how does it arise naturally in the process described in the previous question?
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RESPONSE --> KE is the amount of work that an object is doing at a particular moment in time. If arises naturally from the process described, because if we analyze the units form the derived unit a Joules we see find that Joule = N * m = kg* m/s^2 * m, which shows us that for a given mass we have known acceleration and a known distance, which implies there is an amount of force involved. This gives us the idea that a Joule is simply a unit describing how much force is exerted over a distance, or for what distance at what acceleration a mass is moving. confidence assessment: 3
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21:09:33 ** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. **
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RESPONSE --> Ok self critique assessment: 3
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21:13:16 What forces act on an object as it is sliding up an incline?
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RESPONSE --> We can break this down into external forces acting upon an object, and then of course by Newtons third law the object exerting equal force in the opposite direction of each of these external forces. For an object sliding up a ramp we have the force of gravity acting downwards, the force of friction opposing the direction of the object, and we also have the incline applying an upward force equal and opposite of the gravitational force. confidence assessment: 3
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21:14:42 ** Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force. Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion. The gravitational force is conservative; all other forces in the direction of motion are nonconservative. COMMON ERROR: The Normal Force is in the upward direction and balances the gravitational force. COMMENT: The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal. The normal force cannot balance the gravitational force if the incline isn't horizontal. Friction provides a component parallel to the incline and opposite to the direction of motion. **
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RESPONSE --> Ok, I did not make the qualifcation as I should have that the incline's force that matched gravitational force was only in the perpendicular direction. self critique assessment: 3
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21:16:09 For an object sliding a known distance along an incline how do we calculate the work done on the object by gravity? How do we calculate the work done by the object against gravity?
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RESPONSE --> Fnet * 'ds = 'dW for the work of the object up the incline. For the work against gravity would find the product of Fnet * 'ds * slope of the incline, as long as the incline of the slope was small in value. confidence assessment: 3
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21:16:56 ** The gravitational force is m * g directly downward, where g is the acceleration of gravity. m * g is the weight of the object. If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of which is positive and/or negative. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. The precise value of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal. This force acts down the incline. If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase. This behavior is consistent with our experience of objects moving freely down inclines. If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative. In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines. The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **
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RESPONSE --> Ok self critique assessment: 3
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21:19:38 For an object sliding a known distance along an incline how do we calculate the work done by the object against friction? How does the work done by the net force differ from that done by gravity?
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RESPONSE --> For the work done against friction we take the force of friction or Ffriction * 'ds = 'dW, and apply the correct signs to indicate that the friction opposes the direction of the object. The net force is a force less than gross force due to the opposing frictional force, and gravity will be in value equal to that of the gross force, regardless of the frictional force value. confidence assessment: 3
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21:22:21 ** The work done against friction is the product of the distance moved and the frictional force. Since the force exerted by friction is always opposed to the direction of motion, the force exerted by the system against friction is always in the direction of motion so the work done against friction is positive. The net force on the system is sum of the gravitational component parallel to the incline and the frictional force. The work done by the net force is therefore equal to the work done by gravity plus the work done by the frictional force (in the case of an object moving up an incline, both gravity and friction do negative work so that the object must do positive work to overcome both forces; in the case of an object moving down an incline gravity does positive work on the system while friction, as always, does negative work on the system; in the latter case depending on whether the work done by gravity on the system is greater or less than the frictional work done against the system the net work done on the system may be positive or negative) **
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RESPONSE --> I did not go into detail of how gravity would differ by being able to oppose or assist the motion of the object, and should have since my interpretation only included gravity opposing and not assisting the object's motion along the incline. self critique assessment: 3
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21:27:31 Explain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length.
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RESPONSE --> By definition weight is m * g, which in the case of a pendulum the same value as the Fnet that is the restoring force; therefore, weight of a pendulum is equal to Fnet = m * g = m * g. If the displace was large relative to pendulum length then the relative amount of gravity acting as a restoring force would differ greatly from the force that determines the weight of the pendulum. This is assumption is comparable to the one we make for small slopes on a incline and how much force is caused by gravity opposing an objects motion. confidence assessment: 3
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21:30:11 ** In terms of similar triangles: The reason the approximation only works for small displacements is because the sides used on one triangle are not the same as the sides used on the other. From the triangle we see that the restoring force and the weight are at right angles, while the length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle and hence are not at right angles. For small angles the two long sides of the triangle are approximately equal so the discrepancy doesn't make much difference. For larger angles where the two long sides are significantly different in length, the approximation no longer works so well. In terms of components of the vectors: The tension force is in the direction of the string. The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement (just draw the picture). The vertical component of the tension force must be equal to the weight of the pendulum, since the pendulum is in equilibrium. If the displacement is small compared to the length the vertical component of the tension force will be very nearly equal to the tension force. So the previous statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement' can be replaced by the statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the weight of the pendulum as the length of the pendulum to the horizontal displacement. **
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RESPONSE --> Ok, I explained mine by analogy to slopes of an incline, but please let me know if that is not a proper analogy for this case. self critique assessment: 3
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21:30:18 prin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.
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RESPONSE --> Ok confidence assessment: 3
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21:30:26 The net force on the crate must be zero, since it is not accelerating. The gravitational force on the crate is 160 kg * 9.8 m/s^2 = 1570 N, approx. The only other vertical force is the normal force, which must therefore be equal and opposite to the gravitational force. As it slides across the floor the crate experiences a frictional force, opposite its direction of motion, which is equal to mu * normal force, or .50 * 1570 N = 780 N, approx.. The only other horizontal force is exerted by the movers, and since the net force on the crate is zero the movers must be exerting a force of 780 N in the direction of motion. The work the movers do in 10.3 m is therefore work = Fnet * `ds = 780 N * 10.3 m = 8000 N m = 8000 J, approx..
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RESPONSE --> Ok self critique assessment: 3
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21:30:34 gen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h.
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RESPONSE --> Ok confidence assessment: 3
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21:30:42 To accelerate the helicopter at .10 g it must experience net force Fnet = mass * acceleration = M * .10 g = .10 M g. The forces acting on the helicopter are its upward thrust T and the downward pull - M g of gravity, so the net force is T - M g. Thus we have T - M g = .10 M g, and the upward thrust is T = .10 M g + M g = 1.10 M g. To exert this force through an upward displacement h would therefore require work = force * displacement = 1.10 M g * h = 1.10 M g h.
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RESPONSE --> Ok self critique assessment: 3
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21:52:21 **** Univ: 6.58 (6.50 10th edition). chin-up .40 m, 70 J/kg of muscle mass, % of body mass in pullup muscles of can do just 1. Same info for son whose arms half as long.
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RESPONSE --> For 'dW = Fnet * 'ds, we find that the force would be the man's mass multiplied by the acceleration of gravity; therefore, 'dW/ kg = a * 'ds = 9.8 m/s^2 * .4 m = 3.92 J/ kg of work. If the man the man can do 70 J/ kg of muscle then we find that [(3.92 J/ kg)/ (70 J/ kg)] * 100 = 5.6% of his body is muscle. For his son we find that 'dW/ kg = 9.8 m/s^2 * .2 m = 1.96 J/ kg; therefore, the son has 2.8 % of muscle composition. Even though my answer doesn't support this a child would have should have an easier time doing chin-ups than an adult, because they are have the same muscle to weight ratio, but have to do less work to acheive the same amout numer of chin-ups. confidence assessment: 3
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21:52:40 ** For each kg of mass the weight is 1 kg * 9.8 m/s^2 = 9.8 N. Work done to lift each kg of mass .4 m would then be 9.8 N * .4 m = 3.92 J. The chin-up muscles generate 3.92 J per kg, which is 3.92 / 70 of the work one kg of muscle mass would produce. So the proportion of body mass in the pullup muscles is 3.92 / 70 = .056, or 5.6%. For the son each kg is lifted only half as far so the son only has to do half the work per kg, or 1.96 J per kg. For the son the proportion of muscle mass is therefore only 1.96 / 70 = 2.8%. The son's advantage is the fact that he is lifting his weight half as high, requiring only half the work per kg. **
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RESPONSE --> Ok self critique assessment: 3
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22:08:40 Univ. 6.72 (6.62 10th edition). net force 5 N/m^2 * x^2 at 31 deg to x axis; obj moves along x axis, mass .250 kg, vel at x=1.00 m is 4.00 m/s so what is velocity at x = 1.50 m?
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RESPONSE --> First we calculate what percentage of the force will be applied in the x-direction by the (5 N/m^2)x^2 * cos (31) = (4.29 N/m^2)x^2. Now we can apply the values given to find that (4.29 N/m^2)(1.5)^2 = 4.29 N/m^2 * 2.25 m^2 = 9.6525 N. We can then find a = 9.6525 N/ .25 kg = 38.61 m/s^2. This with a 'ds = 1.5 m - 1 m = .5 m, will allow us to find vf = 'sqrt ( (4.0m/s)^2 + 2 (38.61 m/s^2) * .5 m) = ( 16 m^2/s^2 + 38.61 m^2/s^2) = 7.39 m/s. confidence assessment: 3
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22:09:09 ** Force is variable so you have to integrate force with respect to position. Position is measured along the x axis, so you integrate F(x) = - k / x^2 with respect to x from x1 to x2. An antiderivative of - k / x^2 is k / x so the integral is k / x2 - k / x1. If x2 > x1, then k / x2 < k / x1 and the work is negative. Also, if x2 > x1, then motion is in the positive x direction while F = - k / x^2 is in the negative direction. Force and displacement in opposite directions imply negative work by the force. For slow motion acceleration is negligible so the net force is practically zero. Thus the force exerted by your hand is equal and opposite to the force F = - k / x^2. The work you do is opposite to the work done by the force so will be - (k / x2 - k / x1) = k/x1 - k/x2, which is positive if x2 > x1. This is consistent with the fact that the force you exert is in the opposite direction to the force, therefore in the positive direction, as is the displacement. Note that the work done by the force is equal and opposite to the work done against the force. **
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RESPONSE --> Ok self critique assessment: 3
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