course Phy 231 6/28 1:45 015. `query 15********************************************* ********************************************* Question: `qSet 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? Your solution: You can find the change in momentum. Momentum = Fnet * dt. Confidence rating #$&*: 3
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Given Solution: `a** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qWhat is the definition of the momentum of an object? Your solution: Momentum is found by multiplying the mass of the object by the change in velocity. P = m*v Confidence rating #$&*: 3
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Given Solution: `a** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* 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? Your solution: You multiply the average force by the time interval to find the change in momentum. Confidence rating #$&*: 3
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Given Solution: `a** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law? Your solution: A = F/m Vf = v0+a dt Vf = v0+(F/m)dt Vf-v0 = (F/m)dt Dv = (F/m) dt M dv = F dt confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* 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? Your solution: You can use the v0 and vf to find the acceleration and then use the equation F=ma to find the force. You can also use the change in velocity times the mass to find the momentum, which is also the same as the impulse. Use the number you got and divide it by the time to get the net force on the object. Confidence rating #$&*: 3
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Given Solution: `a** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qClass notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy? Your solution: F=ma => a = F/m Vf^2 = v0^2 +2(F/m)(ds) Vf^2 =v0^2 + 2 F / m ds 1/2 m vf^2 = 1/2 m v0^2 + F ds F ds = (1/2)mvf^2 – (1/2) m v0^2 KE = F * ds KE = work Confidence rating #$&*: 3
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Given Solution: `a** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qWhat is kinetic energy and how does it arise naturally in the process described in the previous question? Your solution: Kinetic energy is (1/2) m*v^2. The change in KE was previously to be the same as the work done by the system: net force times position change. Confidence rating #$&*: 3
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Given Solution: `a** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qWhat forces act on an object as it is sliding up an incline? Your solution: Friction against the direction of motion, gravity (one component is parallel to the direction of motion and one is perpendicular down) and the normal force perpendicular to the incline all work on the object. Confidence rating #$&*: 3
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Given Solution: `a** Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force. Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion. The gravitational force is conservative; all other forces in the direction of motion are nonconservative. More rigorous reasoning: The acceleration of the system is zero in the direction perpendicular to the incline (i.e., the object neither accelerates up and off the incline, nor into the incline). • From this we conclude that the sum of all forces perpendicular to the incline is zero. • In this case the only forces exerted perpendicular to the incline are the perpendicular component of the gravitational force, and the normal force. • We conclude that the sum of these two forces must be zero, so in this case the normal force is equal and opposite to the perpendicular component of the gravitational force. The forces parallel to the incline are the parallel component of the gravitational force and the frictional force; the latter is in the direction opposite the motion of the object along the incline. As the object slides up the incline, the parallel component of the gravitational force and the frictional force both act down the incline. The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal. The normal force cannot balance the gravitational force if the incline isn't horizontal. Friction provides a component parallel to the incline and opposite to the direction of motion. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* 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? Your solution: You calculate the force on the object by multiplying the acceleration due to gravity by the mass of the object to get the net force by gravity. Then you multiply that by the sine of the angle of the incline from horizontal. Finally, you multiply that by distance travelled and you get the work done on the object as it moves down the incline. When the object is going up the incline the work will be negative because the KE is decreasing. The work against gravity is the opposite of the work done by gravity. Confidence rating #$&*: 2
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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 whether the product is positive or negative. • If the displacement `dy is in the same direction as the weight m * g then the product is negative. • If the displacement `dy and the weight m * g are in the same direction then the product is positive. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the magnitude of 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 magnitude 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) ** 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. 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): I’m not really sure if I got this right. I used the work equation but there were things in the explanation that I did not understand. Self-critique rating #$&* 2
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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): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qExplain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length. Your solution: When you draw the picture of the pendulum, the length will be the same as the vertical component of the tension force. Also, the horizontal displacement of the pendulum will be the same as the horizontal component of the tension force. The lengths or forces are proportional only when the displacements are small. When they get larger the proportionality does not hold. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qprin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50. Your solution: 160kg *9.8m/s^2 = 1568 N (this is the gravitational force and the normal force) 1568 N * .5 = 784 N (frictional force) 784N * 10.3m = 8075.2 J of work confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `qgen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h. Your solution: Weight = M*.1g = .1 M g Fnet = thrust – pull of gravity T – Mg = .1Mg T = 1.1Mg Work = 1.1Mg * h = 1.1Mgh Confidence rating #$&*: 2
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* ********************************************* Question: `q**** Univ: 6.58 (6.50 10th edition). chin-up .40 m, 70 J/kg of muscle mass, % of body mass in pullup muscles of can do just 1. Same info for son whose arms half as long. Your solution: Each kg of mass = 1kg *9.8m/s^2 = 9.8N Lift them .4m: 9.8N * .4m = 3.92 J 3.92J / 70J/kg = .056 or 5.6% of muscle mass used 9.8N * .2m = 1.96J 1.96J/70J/kg = .028 or 2.8% of muscle mass used Confidence rating #$&*: 3
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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): OK Self-critique rating #$&* OK Question: (University Physics students only) What is the work done by force F(x) = - k / x^2 between x = x1 and x = x2. Your solution: The force is not constant, so you have to integrate. -k(x^-2)=> –k(x^-1) / -1 = k/x from x1 to x2 = k/x2 – k/x1 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `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): OK Self-critique rating #$&* OK