course PHY 121 1/4 Question: `qSet 4 probs 1-7If 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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Given Solution: ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. ** Self-critique (if necessary): ?? I thought Fnet * `dt was the impulse, and m * v was the momentum? ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qWhat is the definition of the momentum of an object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum = m * v. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ---- ********************************************* 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: Change in momentum = m*`dv. If you know the force and the `dt, which is the impulse, it will be equal to the change in momentum. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I answered my own questions about problem 1.. I understand the quantities are equal, I think the wording just threw me off. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The two equations of the impulse momentum theorem are m*v and fNet * `dt. Uniform accelerated motion = vf = vi + a*`dt, so `dv [the same as (vf – vi)] = a* *`dt. Since a = Fnet/m, `dv = (Fnet/m) * `dt. To make it look like the imp.mom. equations, we have to get m on the other side. So m*`dv = Fnet * `dt. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 ** STUDENT QUESTION: I do not understand why m is multiplied by both sides? INSTRUCTOR RESPONSE: The object is to get new and meaningful quantities on both sides. `dv is nothing new, (F / m) `dt is something we haven't seen before, but is recognizably the same thing as a * `dt. m `dv and F `dt, however, are new concepts. We call mv the momentum, so m `dv is the change in momentum. We call F `dt the impulse. Momentum and impulse are new and useful concepts, a significant addition to our self of 'thinking tools'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ---- ********************************************* 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: If we know m, Vf and Vi, and `dt, we can find the force by multiplying m * `dv and dividing by `dt. Or you could find the acceleration using the velocities and time since a = Fnet / m. Either method results in the Fnet. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 – I really get the impulse/momentum stuff! ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: (Class notes) How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For Sir Newton, Fnet = a * m. Vf^2 = Vi^2 + 2*a*`ds. Our goal is to somehow get to `dWnet = Fnet * `ds on one side, equal to another equation on the other side that relates to energy. If we replace a with Fnet/m, we get vf^2 = vi^2 + 2 (Fnet/m)(`ds). vf^2 (m) = vi^2 + 2*Fnet *`ds. vf^2 (m) – vi^2 = 2 * Fnet * `ds. (vf^2*m – vi^2)/2 = Fnet * `ds. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** STUDENT QUESTION: Why is vf and v0 both ^2 in these equations?? INSTRUCTOR RESPONSE: Note that the units of 2 a `ds are the same as the units for v^2. If the equation had just the first power of v it wouldn't be dimensionally consistent. It takes more than dimensional consistency to make an equation valid, but if the equation isn't dimensionally consistent the equation cannot be valid. The reason v0 and vf are squared: The fourth equation of uniformly accelerated motion is vf^2 = v0^2 + 2 a `ds. This equation was derived earlier in the course; it comes from eliminating `dt between the first two equations. The first two equations come directly from the definitions of velocity and acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought `dKE and `dWnet were being used interchangeably earlier to describe the relationship of Fnet * `ds but want to make sure that’s correct since I haven’t seen `dWnet in a while.
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Given Solution: ** 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): Needed to use the equation 1/2m * v^2. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qWhat forces act on an object as it is sliding up an incline? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As an object slides up an incline, gravity works on the object to “push” it back down and friction from the incline also works on the object. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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): I’m still trying to wrap my head around the two planes of gravity but I think I’m OK. ------------------------------------------------ Self-critique rating: ---- ********************************************* 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 work done on the object is g*m, which is the total gravity on the object. To calculate the work done by the object against gravity, we have to use the displacement as a reference. If the force of the object against gravity is less than the force of gravity, the displacement will be positive since gravity is overpowering the object. But if the force of the object against gravity is greater than the force of gravity on the object, the displacement will be negative as the object “overcomes” gravity. So in this instance I don’t know exactly how to calculate the work done by the object against gravity but we can see whether it is greater or less than that of gravity by comparing it to the displacement. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. (You have seen that the parallel component is m g cos(270 deg - theta) or m g cos(270 deg + theta), depending on whether your incline slopes up or down as you go left to right. These expressions follow directly from the circular definition of the trigonometric functions. The magnitude of cos(270 deg - theta) is the same as the magnitude of cos(270 deg + theta), and each is in turn the same as the magnitude of sin(theta). The expression m g * sin(theta) also follows directly from the right-angle trigonometry of the situation.) 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): I’m getting quite confused about all the theta, cos, sin stuff.. I feel like I must have skipped over some notes or something because I am drawing a complete blank. ------------------------------------------------ Self-critique rating: ---- ********************************************* 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: The distance the object travels is relative to the work done against friction, combating the friction. The force of gravity involves gravity and the mass of the object, while the Fnet includes both the friction and the force of gravity. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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, but I need to look over again. ------------------------------------------------ Self-critique rating: ---- ********************************************* 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** The vertical leg and hypotenuse are of very nearly the same length. Therefore the vertical component of the tension is very nearly equal in magnitude to the tension. As the pendulum swings back toward equilibrium it moves almost entirely in the horizontal direction, and therefore has practically no vertical acceleration. The net vertical force is therefore practically zero. We conclude that the vertical component of the tension is very nearly equal and opposite the weight of the pendulum. In terms of the notation of the next figure, in which the sides of the triangle are labeled in terms of the tension and its components, we see that T_y is equal to the weight of the pendulum T is very nearly the same as T_y T_x is the horizontal component of the tension. • In this figure T_x appears to be about 1/10 as great as T_y. • So we say that the ratio T_x / T_y is roughly 1/10, or .1. Since T_y is equal to the weight of the pendulum, T_x / T_y is the ratio of the x component of the tension to its weight: • T_x / T_y = T_x / weight In the next figure we superimpose a similar triangle on the original sketch of the pendulum. The length of the triangle is equal to the length of the pendulum, and the horizontal leg is the displacement of the pendulum from its equilibrium position. Sketching the triangle by itself and labeling its hypotenuse L (for the length of the pendulum) and its horizontal leg x we have the figure below The vertical leg of this triangle is very nearly the same length as L. So the ratio of horizontal to vertical legs is very close to x / L. This triangle is geometrically similar to the triangle we used previously to represent the components of the tension. The geometric similarity implies that the ratio of horizontal to vertical leg must be the same for both. Writing this condition in symbols we have T_x / T_y = x / L Since T_y is nearly the same as the weight we have T_x / weight = x / L. That is, the force restoring the pendulum to equilibrium is in the same proportion to the weight of the pendulum as the displacement from equilibrium to its length. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qprin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 160kg `ds = 10.3m. friction = .5. Fnet * `ds = `dWnet, but we don’t know Fnet. Fnet = m * g = 160kg *9.8m/s^2 = 1568 N. If friction = .5, then friction = .5(1568N) = 784N. 1568-784 = 784N. 784 * 10.3m = work done = 8075 J. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK – though I had to make some assumptions. I don’t know why the questions keep getting all jumbled. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: What is the work done by force F(x) = - k / x^2 between x = x1 and x = x2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have no idea how to answer this question. All I can come up with is that we’re looking for the work done by the force between x1 and x2, but I’m not sure whether x1 and x2 are considered distance or velocity. The only thing I can think of is that since we know Fnet and are looking for `dWnet, we need to use Fnet * `ds = `dWnet. That would mean (-k/x^2) * (x2-x1) = `dWnet. So (-k/x^2) * x1 = `dWnet = -k/x. Confidence rating: 1 ** 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): I think I was on the right track to begin with, but I don’t understand why we use x, x1, and x2, instead of just x1 and x2. ------------------------------------------------ Self-critique rating: "