course Phy201 Query 15 part 2 015. `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): ok ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qWhat is the definition of the momentum of an object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Definition of momentum of an object is the net force on the object times the amount of time confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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, it is mass times the velocity i had the impulse. ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qHow do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would multiply the time interval times the average force acting on the object Which is the impluse, since imuplse= momentum confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: impulse = fnet * 'dt change in momentum= m * v Fnet can be found by mass * acceleration. Velocity can be found by KE= .5m*v^2 If you know the velocity and acceleration you can obtain the time. Once you have the Fnet and the time you can figure the impulse, and if you have the mass of the object times the velocity you have the momentum. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ** 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 I understand this ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qIf you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we know mass v0 vf 'dt need to find fave you can find acceleration by a= 'dv/'dt. Once you find a you can get force by f= m* a or use the momentum theory change in momentum = m* v by mulitplying the change in velocity by the mass. Since momentum = change in impulse and change in impulse = fnet * 'dt you would get fnet by dividing impluse by the 'dt ( fnet= impulse/'dt). confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 I understand this ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qClass notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: newtons second lw states that acceleration is proportional to the net force actin on the object and is inversely proportional to its mass. Direction of acceleratin is the direction of net force acting on the object. KE= .5m v^2 so kinetic energy = half the mass times the square of the veolcity. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** 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): ok I think that i get this, or at least I see how they relate. ------------------------------------------------ Self-critique rating #$&*2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qWhat is kinetic energy and how does it arise naturally in the process described in the previous question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE= .5m V^2 which is also = to the 'dw of an object. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 I understand ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qWhat forces act on an object as it is sliding up an incline? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The forces that act on an object as it is sliding up an incline are the frictioanl forces of the object against the surface of the incline that are doing work on the object and the forces that are being done by the object as is goes up the incline. THe work that is being done by the oject as it goes up the inlcine are opposing the work done on the object by friction as it goes up the incline. So there are forces that are from work done by the object and forces that are from work being done on the object by friction. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok I got most of this, I think I am getting it. ------------------------------------------------ Self-critique rating #$&*2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qFor an object sliding a known distance along an incline how do we calculate the work done on the object by gravity? How do we calculate the work done by the object against gravity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: known 'ds a= 9.8m/s^2 'dw= fnet * 'ds fnet= m * a To get the work done on the object by gravity we would calculate the force= mass * acceleration. accleration would be 9.8m/s^2 and we would need to know the mass of the object. from there since we have the distance traveled we can obtain the work done by gravity by multiply the fnet time the knwn 'ds traveled. Work done byu the object against gravity would be opposing the forces from the work being done by the object by gravity. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. (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): ok ------------------------------------------------ Self-critique rating #$&*2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qFor an object sliding a known distance along an incline how do we calculate the work done by the object against friction? How does the work done by the net force differ from that done by gravity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'dw = fnet * 'ds we would calculate the frictional force minus the force of the systme moving down the inlcine at m*g*sin(theta) the friction would be opposing this. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) ** STUDENT QUESTION Oops! I forgot that the frictional force is added to the force of the object. But I am a little hazy on the ‘why’ of this. Since friction works against the force of the object, wouldn’t it be subtracted from the Fnet. OR… is the friction added to the Fnet because the object has to the work of itself + the frictional component to move along the incline?? INSTRUCTOR RESPONSE Friction opposes the relative motion of two surfaces. In this case the relative motion is that of the object sliding on the surface of the incline. So the frictional force acts in the direction opposite the motion. The gravitational force component parallel to the incline acts down the incline. So if the motion is up the incline the frictional force is in the same direction as the gravitational force. If motion is down the incline, the frictional force acts in the direction opposite the parallel gravitational component. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I see why you add friction instead of subtracting since friction is going in the direction of motion up the incline. ------------------------------------------------ Self-critique rating #$&*2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qExplain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation would be m*gcos(theta) force of gravity is acting on the weight of the pendulum. The weight plus gravity will determine how fart he pendulum will swing from extreme point to equilibrium. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** This situation is illustrated in the figures below. Note that this is also explained in video embedded in the DVD version of the Class Notes. The first figure depicts a simple pendulum consisting of a symmetric mass suspended from a fixed point by a light string. The pendulum is not in its vertical position, but is displaced a bit to the right of that position. The string exerts a force that prevents the pendulum from accelerating vertically downward. This force results when the string is stretched slightly in response to the weight of the pendulum, and is called a tension force. It acts along the line of the string, pulling the pendulum up and toward the left. The weight of the pendulum is the force exerted on it by gravity. The weight acts in the vertical downward direction. The figure below depicts the tension and the weight. The next figure depicts only the tension force. In the following picture we superimpose a triangle on the preceding picture. The hypotenuse of the triangle coincides with the tension vector. The legs of the triangle are in the horizontal and vertical directions. The triangle is shown by itself below. The length of the hypotenuse represents magnitude of the tension force, the lengths of the vertical and horizontal legs represent the vertical and horizontal components of the tension vector. 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 #$&* ********************************************* Question: `qprin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m= 160kg 'ds= 10.3m mv= .50 'dw= fnet * 'ds f= m* a f= 160 * 9.8m/s^2 = 1568 'dw= 568 * 10.3m =16150.4 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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, mV is the force opposing direction of motion .50 * 1568 = 784 since there is no acceleration thisis the only force acting on the object, so 'dw= 784N * 10.3m = 8075J ------------------------------------------------ Self-critique rating #$&*3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `qgen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not sure about this confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION I totally spaced on the Thrust aspect of this problem. What indicators are there in a problem that tell me to incorporate Thrust, when it doesn’t state it within the problem. Do I have to use it in all problems when things are lifted off the ground?? INSTRUCTOR RESPONSE If it's accelerating upward at .10 g, then the net force is .10 M g. It doesn't matter what 'it' is; could be a helicopter, an elevator, a fish on a line. Since gravity is pulling it downward something, whatever we wish to call it, is pulling it upward, with a force we might as well represent by the letter T (any other letter would do as well; we could revert to the old standby and call the force x). It doesn't hurt to have a word for this force, but we don't really need one; all we really need is to know it's there. It's not necessary to call this force 'thrust' (applicable to the helicopter) or 'tension' (applicable to the elevator or the fish, which are being pulled upward by the tension in a cable or fishing line). But whatever we want to call it (if anything), or whatever symbol we want to use for it, we know it has to be there because if it wasn't the object would be accelerating downward at 9.8 m/s^2. Using T for the unknown force, the net force on the object is T - M g. This gives us two expressions for the net force. We set these two expressions equal and get the equation T - M g = .10 M g and solve to get T = 1.10 M g &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): T = .10 M g + M g = 1.10 M g. I am not understanding how exactly we get 1.10 here??????????????????????????????? How do we calculate this out????????????????????