course Phy 201 015. `query 15
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Given Solution: `a** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qWhat is the definition of the momentum of an object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: momentum = mass * velocity. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qHow 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: impulse = Fave * `dt = change in momentum We can multiply Fave * `dt to get change in momentum. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F=ma, so we know a=F/m vf= v0 + a*`dt, subtract v0: vf-v0 = a*`dt. Since vf-v0 = `dv, This shows us that `dv = a'dt. Substitute a=F/m, we then get `dv = (F/m)*`dt. Then multiply both sides by m, M*`dv = F*`dt Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Impulse-momentum theorem for constant masses states: m `dv = Fave `dt Therefore we can determine, Fave = m `dv / `dt. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qClass notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = F / m. vf^2 = v0^2 + 2 a `ds 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 Therefore, Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2 This is the same as saying work = change in KE. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qWhat is kinetic energy and how does it arise naturally in the process described in the previous question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE = 1/2 m v^2, Which we know to be equal to the work done by the net force. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qWhat forces act on an object as it is sliding up an incline? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Gravitational force has 2 components: one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and 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 motionwould be nonconservative. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. More rigorous reasoning: The acceleration of the system is zero in the direction perpendicular to the incline (i.e., the object neither accelerates up and off the incline, nor into the incline). From this we conclude that the sum of all forces perpendicular to the incline is zero. In this case the only forces exerted perpendicular to the incline are the perpendicular component of the gravitational force, and the normal force. We conclude that the sum of these two forces must be zero, so in this case the normal force is equal and opposite to the perpendicular component of the gravitational force. The forces parallel to the incline are the parallel component of the gravitational force and the frictional force; the latter is in the direction opposite the motion of the object along the incline. As the object slides up the incline, the parallel component of the gravitational force and the frictional force both act down the incline. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qFor 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The gravitational force = m * g going directly downward, g is the acceleration of gravity. m * g = weight of the object. If we know `ds then we can simply multiply weight m * g with the vertical displacement `dy, making sure to note which is negative and which is positive. For small inclines the parallel component of the gravitational force is approximately equal to the product of the weight and the slope of the incline. 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 is positive and, without other forces in this direction, the the object’s KE will increase. This is uniform with objects down an incline. If displacement is upward along the incline then `ds would be in the opposite direction to the gravitational force and the work done by gravity is negative. Without other forces in the direction of the incline there will be a loss of KE. This is uniform with objects coasting up inclines. Work against gravity is opposite (thus negative) of the work done by gravity, and therefore positive for an object moving up an incline and negative for an object moving down the incline (because it would tend to gain energy instead of expending any in this case). Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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) ** NOTE ON THE EXPRESSION m g * sin(theta) 'down the incline' Suppose the incline is at angle theta with horizontal, with the incline ascending as we move to the right. If the x and y axes are in their traditional horizontal and vertical orientations, then the incline makes angle theta with the positive x axis, and the weight vector acts along the negative y axis. It is more convenient to have the x axis directed along the incline, so that motion is along a single axis. We therefore rotate the coordinate system counterclockwise through angle theta, bringing the x axis into the desired alignment. As we do this, the y axis also rotates through angle theta, so that the negative y axis rotates away from the weight vector. When we have completed the rotation, the weight vector will lie in the third quadrant, making angle theta with respect to the negative y axis. The direction of the weight vector will then be 270 deg - theta, as measured counterclockwise from the positive x axis. The x and y components of the weight vector will then be ( m g * cos(270 deg - theta) ) and ( m g * sin(270 deg - theta) ). It turns out that cos(270 deg- theta) = -sin(theta), and sin(270 deg - theta) = -cos(theta), so the x component of the gravitational force is -m g sin(theta); alternatively we can express this as m g sin(theta) directed down the incline. This agrees with the given formula. A displacement `ds up the incline (in the direction opposite the gravitational force component along the incline) implies that work `dW = -m g sin(theta) * `ds is done on the object by gravity, so that its gravitational PE increases by amount m g sin(theta) * `ds. NOTE ON m g sin(theta) * `ds For the same incline as discussed in the previous note, if the displacement is `ds up the incline, then the displacement vector will have magnitude `ds and will make angle theta with the horizontal. If our x and y axes are respectively horizontal and vertical, then the displacement is represented by the vector with magnitude `ds and angle theta. The horizontal and vertical components of this vector are respectively `ds cos(theta) and `ds sin(theta). In particular an object which undergoes displacement `ds up the incline has a vertical, or y displacement `dy = `ds sin(theta). This displacement is along the same line as the gravitational force m g, but in the opposite direction, so that the work done on the object by gravity is - m g * `ds sin(theta), and the change in gravitational PE is again found to be m g sin(theta) * `ds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qFor 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `a** The product of `ds and frictional force equals the work that is being done against friction. Because frictional force is always opposite the direction of motion, the force exerted by the system will in the direction of position therefore the work being done against friction will be positive. Fnet is equal to the frictional force plus the gravitational component parallel to the incline. Work done by the Fnet is therefore equal to the work done by gravity plus the work done by the frictional force. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qExplain 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can this of this in terms of triangles. The approximation only works for small displacements because the sides are not used on both triangles. Restoring force and the weight are at right angles. The length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle are not at right angles. When thinking of small angels, the 2 long sides of the triangle are approx equal so there wont be much difference caused by discrepancy. For larger angles, where there is a significant difference in the length of the 2 long sides, the approximation no longer works very well. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qprin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet on the crate has to be 0, since its not accelerating. The Gravitational force on the crate = 160 kg * 9.8 m/s^2 = approx 1570 N Vertical force which is the normal force is equal and opposite to the gravitational force. As it slides across the floor, friction acts opposite to its direction of motion. Frictional force = mu * normal force = .50 *1570 N = approx 780N The only other horizontal force is exerted by the movers, and since Fnet on the crate is 0, 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 = approx. 8000 J Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `qgen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `aTo accelerate the helicopter at .10 g: 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 gravit. Fnet = T - M g. Thus we have T - M g = .10 M g, and the upward thrust: 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. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aTo 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. STUDENT COMMENT AND INSTRUCTOR RESPONSE: I didn't think of that. I still don't fully understand it. INSTRUCTOR RESPONSE: F_net = m a = m * .10 g = .10 m g. F_net = upward thrust + gravitational force = T - m g. Thus T - m g = .10 m g. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: `q**** 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: (NOTE: THIS PROBLEM HAS BEEN OMITTED FROM THE TEXT AND IS NOT PRESENTLY ASSIGNED) `q Univ. 6.72 (6.62 10th edition). A net force of 5.00 N m^2 / x^2 is directed at 31 degrees relative to the 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: the component of the force in the direction of motion is 5.00 N / m^2 * x^2 * cos(31 deg) = 5.00 N / m^2 * x^2 * .86 = 4.3 N/m^2 * x^2. Integrating this with respect to x from x = 1.00 m to x = 1.50 m we get something around 3 Joules (antiderivative is 4.3 N / m^2 * x^3 / 3 = 1.43 N/m^2 * x^3; the change in the antiderivative is about 1.43 N/m^2 [ ( 1.50 m)^3 - (1.00 m)^3 ] = about 3 N * m = 3 J). Initial KE is 1/2 * .250 kg * (4.00 m/s)^2 = 2 J Final KE is 1/2 * .250 kg * (4.00 m/s)^2 + 3 J = 5 J, approx. so final vel is vf = sqrt( 2 KEf / m) = sqrt( 2 * 5 J / (.250 kg) ) = sqrt( 40 m^2 / s^2) = 6.4 m/s, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment ********************************************* Question: What is the work done by force F(x) = - k / x^2 between x = x1 and x = x2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Assessment Calculus-related comments and observations (applicable mainly to University Physics; may be of interest to students in other physics courses who know calculus): Good observation by student: the derivative of change in KE is mvf-mv0. so i suppose this is the change in momentum...? Instructor response: Good observation. It's important to be very specific about derivatives (e.g., to specify the derivative of what with respect to what): Strictly speaking, for a given interval 1/2 m vf^2 - 1/2 m v0^2 is a fixed quantity, so its derivative is zero. However KE typically changes from one instant to another, and can be regarded as a continuous function of velocity. Specifically KE = 1 / 2 m v^2, where v can vary, so the derivative of this expression with respect to v is m v: d(KE)/dv = m v = momentum In words, the derivative of KE with respect to velocity is momentum. Thus momentum is the rate of change of KE with respect to v. We can go even further: Of course velocity can be regarded as a function of clock time. The chain rule tells us that d(KE) / dt = d(KE) / dv * dv / dt. As we have seen, d(KE)/dv = m v = momentum; and we know that dv/dt = a, the acceleration. So using a(t) to explicitly express the acceleration as a function of clock time, we have d(KE) / dt = (m v) * a(t) = momentum * acceleration. The derivative of the KE with respect to clock time is the product of momentum and acceleration. Integrating the force function for a pendulum or a spring: The force restoring a pendulum to equilibrium (provided displacement from equilibrium is small compared to length), or to restore a spring to its equilibrium position, is of the form F = - k x. This force is conservative in the ideal case. The work done by this force is found by integrating the force with respect to position x. Integrating with respect to x, the force constant k is a constant. The force function is F(x) = - k x; integrating this function with respect to x we get - k x^2 / 2 + c, where c is an integration constant. If x = 0 is taken as the 0-energy point (appropriate to the equilibrium point) then -k * 0^2 / 2 + c = 0, so c = 0 and the work done by the system against the conservative force is - k x^2 / 2. The potential energy being the negative of the work done against the conservative force is PE = k x^2 / 2.