#$&* course Phy 241 October 15 around 3:50pm 015. `query 15
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Given Solution: `a** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qWhat is the definition of the momentum of an object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum is mass*velocity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. ** STUDENT QUESTION ok, I am confused what 'dp stands for in your explanation. INSTRUCTOR RESPONSE p is the standard symbol for momentum. Thus `dp is the change in momentum. `dp = `d( m v ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Favg*’dt=change in momentum (and impulse) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Newton’s Second Law: F=ma or a=F/m Uniformly accelerated motion equation: vf=v0+a’dt Vf-v0=a’dt Change in v=a’dt Change in v=(F/m)’dt m’dv=F’dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt ** STUDENT QUESTION I do not understand why m is multiplied by both sides? INSTRUCTOR RESPONSE The object is to get new and meaningful quantities on both sides. `dv is nothing new, (F / m) `dt is something we haven't seen before, but is recognizably the same thing as a * `dt. m `dv and F `dt, however, are new concepts. We call mv the momentum, so m `dv is the change in momentum. We call F `dt the impulse. Momentum and impulse are new and useful concepts, a significant addition to our self of 'thinking tools'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: One way: (going back to the previous answer) m’dv=Favg*’dt Favg=(m’dv)/(‘dt) Another way: Aavg=(vf-v0)/’dt =(’dv/’dt)*(mass) Will give you Favg. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Newton’s Second Law: F=ma or a=F/m Use: vf^2=v0^2+2a’ds =Vf^2=v0^2+2(F/m)’ds =Vf^2-v0^2=2(F/m)’ds =(vf^2-v0^2)/(2/m)=F’ds = (1/2mvf^2)-(1/2mv0^2)=F’ds F’ds is work (1/2mvf^2)-(1/2mv0^2) is change in KE confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. ** STUDENT QUESTION: Why is vf and v0 both ^2 in these equations?? INSTRUCTOR RESPONSE: Note that the units of 2 a `ds are the same as the units for v^2. If the equation had just the first power of v it wouldn't be dimensionally consistent. It takes more than dimensional consistency to make an equation valid, but if the equation isn't dimensionally consistent the equation cannot be valid. The reason v0 and vf are squared: The fourth equation of uniformly accelerated motion is vf^2 = v0^2 + 2 a `ds. This equation was derived earlier in the course; it comes from eliminating `dt between the first two equations. The first two equations come directly from the definitions of velocity and acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qWhat is kinetic energy and how does it arise naturally in the process described in the previous question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE= ½ mv^2 Previous answer was (1/2mvf^2)-(1/2mv0^2) which is change in KE and equals the work done. 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 But, the work done was BY THE NET FORCE!
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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 need to remember that the normal force balances the grav. force.
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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. FOR THE PERPLEXED If you don't understand this problem or the given solution, you should first be sure you have viewed all assigned Class Notes to date. Then you should specifically self-critique your understanding of the following summary by copying it into your self-critique. Alternatively you could submit these questions and your responses separately, using the Submit Work Form: • #$*& A displacement vector of length `ds in direction theta relative to the positive x axis has components `ds * cos(theta) and `ds * sin(theta). && • #$*& The vertical component of the displacement vector `ds is `ds * sin(theta). && • #$*& If the x-y coordinate system is in the 'standard' orientation, with the x axis horizontal and the y axis vertical, then the weight vector is directed along the negative y axis. && • #$*& If the coordinate system is rotated so that the x axis is directed at angle theta with above horizontal, the displacement vector whose length is `ds and whose direction makes angle theta with respect to horizontal will now point in the x direction. && • #$*& The weight vector lies at angle 270 degrees - theta with respect to the x axis of the rotated system. && • #$*& The components of the weight vector in the x and y direction are, respectively, m g cos(270 deg - theta), and m g sin(270 deg - theta). && • #$*& The displacement vector `ds and the weight component m g cos(270 deg - theta) are both directed along the x axis of the rotated system, the first in the positive direction and the second in the negative. && • #$*& When these two components are multiplied, we get the work done by the gravitational force on the mass. && • #$*& The vertical component `ds * sin(theta) of the displacement vector and the gravitational force m g are both in the vertical direction, one acting upward and the other downward. So when we multiply `ds sin(theta) by (- m g) we get the work done by the gravitational force on the mass. && • #$*& Whichever way we choose to orient our system, we find that the work done by the gravitational force is - m g * `ds * sin(theta). && #$*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got the 2 main concepts. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q For an object sliding a known distance along an incline how do we calculate the work done by the object against a given frictional force? How does the work done by the net force differ from that done by gravity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Work done by the object against a given frictional force: (‘ds*Ffric) The work done against frictional force is always POSITIVE since “Ffric” is opposite to the direction of motion, while the force BY the system IS in the direction of motion. Net force vs. Gravity (work by Fnet=work(grav)+work(Ffric)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 could have went more in detail with the net force vs. gravity. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qExplain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You have to look at “tension force” on the object to answer these. The tension force is in the direction of the string. The component of tension in the horiz. direction=total tension as length of pendulum to the horiz. displacement. The vertical tension component=weight of pendulum (due to equilibrium). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Note that this topic was covered in Class Notes #14, which are part of the preceding assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): A little confusing, but I got the main idea down. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q**** Univ: 6.58 (6.50 10th edition). chin-up .40 m, 70 J/kg of muscle mass, % of body mass in pullup muscles of can do just 1. Same info for son whose arms half as long. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1kg*9.807m/s^2=9.807N 0.4m*9.807N=3.9228J 3.9228J/(70J/kg)=0.5604 (5.604%) Arms half as long: 3.9228J/2=1.9614J 1.9614J/(70J/kg)=0.2802 (2.802%) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): PERFECT! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: (NOTE: THIS PROBLEM HAS BEEN OMITTED FROM THE TEXT AND IS NOT PRESENTLY ASSIGNED) `q Univ. 6.72 (6.62 10th edition). A net force of 5.00 N m^2 / x^2 is directed at 31 degrees relative to the x axis. ; obj moves along x axis, mass .250 kg, vel at x=1.00 m is 4.00 m/s so what is velocity at x = 1.50 m? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5*x^2*cos (31)=4.3x^2 Integrate (4.3x^2) from 1 to 1.5 =3.4J KE=1/2mv^2 Starting KE=1/2(0.250)(4^2)=2J Ending KE=2J+3.4J=5.4J Vf=sqrt ((2Kef)/m) =6.57m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Rounding and sig figs made a slight difference in solutions. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: (University Physics students only) What is the work done by force F(x) = - k / x^2 between x = x1 and x = x2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m somewhat confused, but I’ll attempt, using what is giving and what I know: Work BY is equal and opposite to work AGAINST If x2>x1 then work would be NEGATIVE since ((k/x2)<(k/x1)), this shows the work done BY the system Work done by me is equal and opposite to work done BY force, which would be POSITIVE. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `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 It definitely made better sense when I read your solution.