course Phy 121
.................................................
......!!!!!!!!...................................
21:01:04 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
......!!!!!!!!...................................
RESPONSE --> a= Fnet / mass = 10 Newtons / (2 kg) = 5 m/s^2 `dv = 5 m/s^2 * 3 s = 15 m/s
.................................................
......!!!!!!!!...................................
21:01:07 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:04:08 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
......!!!!!!!!...................................
RESPONSE --> 2 kg * 15 meters/second = 30 kg m/s Fnet *`dt = 10 Newtons * 3 seconds = 30 kg m/s They are the same.
.................................................
......!!!!!!!!...................................
21:04:15 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:07:12 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
......!!!!!!!!...................................
RESPONSE --> Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 kg meters/second
.................................................
......!!!!!!!!...................................
21:07:16 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:09:33 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
......!!!!!!!!...................................
RESPONSE --> 1200 kg * 1.5 s = 3000 kg m/s `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s
.................................................
......!!!!!!!!...................................
21:09:37 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:14:23 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
......!!!!!!!!...................................
RESPONSE --> m * `dv = 1600 kg * 5 meters/second = 8000 kg m/s Fnet * 2 sec = 8000 kg m/s Fnet = 8000 kg m/s / 2s = 4000 kg m/s^2 = 4000 Newtons
.................................................
......!!!!!!!!...................................
21:14:27 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
ėˠEîVJd assignment #015 [뺗܃ܝ Physics I 02-28-2006
......!!!!!!!!...................................
23:34:34 Set 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?
......!!!!!!!!...................................
RESPONSE --> Momentum Fnet * 'dt = momentum
.................................................
......!!!!!!!!...................................
23:34:39 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:35:58 What is the definition of the momentum of an object?
......!!!!!!!!...................................
RESPONSE --> Momentum is the product of the mass and velocity.
.................................................
......!!!!!!!!...................................
23:36:27 02-28-2006 23:36:27 ** 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. **
......!!!!!!!!...................................
NOTES -------> Change in momentum is mass * change in velocity assuming constant mass
.......................................................!!!!!!!!........................... ........
23:39:09 How 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?
......!!!!!!!!...................................
RESPONSE --> 'momentum = Fave * 'dt
.................................................
......!!!!!!!!...................................
23:39:15 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:43:52 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
......!!!!!!!!...................................
RESPONSE --> If F=ma then a=F/m and if vf= v0 + a'dt then vf-v0 = a'dt which becomes 'dv = a'dt 'dv = (F/m)'dt and so m'dv = F'dt
.................................................
......!!!!!!!!...................................
08:44:06 ** 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 **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:47:08 If 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?
......!!!!!!!!...................................
RESPONSE --> impulse-momentum theorem for constant masses is m `dv = Fave `dt and so Fave = m `dv / `dt aAve = (vf - v0) / `dt times the constant mass to get Fave
.................................................
......!!!!!!!!...................................
08:47:23 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:49:55 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
......!!!!!!!!...................................
RESPONSE --> a = F / m vf^2 = v0^2 + 2 a `ds vf^2 = v0^2 + 2 (Fnet / m) `ds Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2
.................................................
......!!!!!!!!...................................
08:50:02 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:54:19 ** 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. **
......!!!!!!!!...................................
RESPONSE --> KE is 1/2 m v^2 the change in KE is = to the work done by net force
.................................................
......!!!!!!!!...................................
08:55:21 What forces act on an object as it is sliding up an incline?
......!!!!!!!!...................................
RESPONSE --> gravity and friction
.................................................
......!!!!!!!!...................................
08:56:03 ** 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. 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. **
......!!!!!!!!...................................
RESPONSE --> gravity parallel and perpendicular to the ramp
.................................................
......!!!!!!!!...................................
09:02:25 For 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?
......!!!!!!!!...................................
RESPONSE --> I meant to save the last comment as notes. I don't think I really understand this concept very well.
.................................................
......!!!!!!!!...................................
09:03:31 ** The gravitational force is m * g directly downward, where g is the acceleration of gravity. m * g is the weight of the object. If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of which is positive and/or negative. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. The precise value of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal. This force acts down the incline. If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase. This behavior is consistent with our experience of objects moving freely down inclines. If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative. In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines. The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **
......!!!!!!!!...................................
RESPONSE --> I just copy and pasted the answer for future reference but could you try to explain this in another way?
.................................................
......!!!!!!!!...................................
09:10:28 For 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?
......!!!!!!!!...................................
RESPONSE --> Work done against friction = distance moved * frictional force 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 equal to the work done by gravity plus the work done by the frictional force.
.................................................
......!!!!!!!!...................................
09:10:37 ** 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) **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
12:24:26 Explain 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.
......!!!!!!!!...................................
RESPONSE --> I don't know the answer to this question I think this comes from the experiment 7 and I did not have the materials to that experiment.
.................................................
......!!!!!!!!...................................
12:24:41 03-01-2006 12:24:41 ** 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. **
......!!!!!!!!...................................
NOTES -------> 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
.......................................................!!!!!!!!........................... ........
12:27:14 prin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.
......!!!!!!!!...................................
RESPONSE --> gravitational force on the crate is 160 kg * 9.8 m/s^2 = 1570 N .50 * 1570 N = 780 N 780 N * 10.3 m = 8000 N m = 8000 J
.................................................
......!!!!!!!!...................................
12:27:18 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..
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
12:27:24 gen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h.
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
......!!!!!!!!...................................
12:27:29 To 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.
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
......!!!!!!!!...................................
12:27:34 **** 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.
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
......!!!!!!!!...................................
12:27:39 ** 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. **
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
......!!!!!!!!...................................
12:27:44 Univ. 6.72 (6.62 10th edition). net force 5 N/m^2 * x^2 at 31 deg to 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?
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
......!!!!!!!!...................................
12:27:49 ** 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. **
......!!!!!!!!...................................
RESPONSE --> N/A
.................................................
"
Good work on most questions. See my note and let me know if you have further questions.