course PHY231 ??????h-??|???R?assignment #015
......!!!!!!!!...................................
13:52:16 `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 --> Acceleration is 5 m/sec^2 using a = Fnet/m Acceleration of 5 m/sec^2 for 3 seconds results in change in velocity of 15 m/sec confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:52:21 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:00:20 `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 --> The quantity m * v changed by 30 kg* m/sec during the three seconds. The quantity Fnet 'dt also changed by 30 kg* m/sec during the three seconds. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:00:31 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:05:06 `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 --> The impulse of the 2 kN average net force applied over 1.5 seconds is Fnet 'dt = 3 kN confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:06:16 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 --> I wrote the units correctly, but typed them in incorrectly . . . self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:10:05 `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 --> The quantity mv will change by the impulse calculated previously (3000 kg * m/sec). Dividing that quantity by the mass of the vehicle will give the change in velocity: (3000 kg * m/sec)/ 1200 kg = 2.5 m/sec confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:10:30 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:15:15 `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 --> Solving the previous equation for Fnet, we get: ('dv * m) / 'dt = Fnet Substituting 'dv = +5 m/sec, 'dt = 2 sec, m = 1600 kg We solve to get 4 kN confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:15:21 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 --> self critique assessment: 3
.................................................
??H??|???~?????? assignment #015 015. `query 15 Physics I 06-22-2007
......!!!!!!!!...................................
15:28:15 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 --> The product of Fnet 'dt is the impulse of the force, which is a quantity that is equal to the change in momentum. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:28:40 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
15:30:28 What is the definition of the momentum of an object?
......!!!!!!!!...................................
RESPONSE --> Momentum is the measure of the effectiveness of an object in a collision. It is the product of mass and velocity. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:31:32 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:32:44 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 --> The change in momentum is equal to the impulse, which is the product of Fave and 'dt. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:33:00 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:40:28 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
......!!!!!!!!...................................
RESPONSE --> Newton's second law can be stated: a = F/m The equation of motion we will use is vf^2 = v0^2 + a * 'dt Substituting F/m for a, we can rearrange the equation to (vf^2 - v0^2) * m = F 'dt. This states that the product of mass and the change in velocity is equal to the product of the force and time interval. Since the product mv equals momentum and the product F 'dt equals impulse, this equation states that the impulse of a force is equal to the momentum of an object. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:42:12 ** 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 --> I dropped the change from my velocity in the middle of that. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:49:33 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 --> With the final and initial velocities, we can easily calculate the change in velocity. We've then got the product m 'dv = p. Knowing that impulse is equal to momentum, we can solve for Fave = (m * 'dv) / 'dt. We can also calculate the acceleration, 'dv /'dt and multiply by the mass to find F = m*a. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:49:53 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:56:28 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
......!!!!!!!!...................................
RESPONSE --> We can write Newton's second law as a = F/m Using the equation vf^2 = v0^2 + 2 a 'ds, we substitute F/m for a and rearrange: vf^2 - v0^2 = 2/m F 'ds , then: m/2 (vf^2 - v0^2) = F 'ds confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:56:52 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:00:43 What is kinetic energy and how does it arise naturally in the process described in the previous question?
......!!!!!!!!...................................
RESPONSE --> Kinetic energy is the energy of motion and is expressed: m/2 (v^2). In the previous question, the change in kinetic energy was equal to the amount of work done when the net force acted over a specific displacement. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:00:51 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:05:52 What forces act on an object as it is sliding up an incline?
......!!!!!!!!...................................
RESPONSE --> As an object slides up an incline, there is a component of gravitational force (parallel to the incline) that is not balanced by the normal force (perpendicular to the incline). This component of gravitational force and a frictional force that opposes motion are the forces that act on the object. confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:07:10 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:28:13 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 --> The work done on any object by the gravitational force is always the product of the object's weight (w = mg) and its vertical displacement. Knowing the displacement along the incline alone is not enough. We can find the vertical displacement if we know the angle of the incline (theta), by calculating 'ds * sin(theta) = vertical displacement. If we're given horizontal displacement as well as displacement along the incline, we can use the pythagorean theorem to find the vertical displacement. If we're given the slope of the incline, and the slope is small, we can approximate the gravitational force parallel to the incline by the product of the object's weight and the slope of the incline. In this case, we us the given value of displacement along the incline. Gravity does positive work on an object as it slides down an incline, as the force and motion are in the same direction. The negative of that work is the work done by the object against gravity. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:29:03 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:44:54 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 --> If we are given the mass and change in velocity of the object, we can calculate it's kinetic energy (m/2 * v^2). This is the work done on it by the net force. If gravity and friction are the only forces acting on the object, the work done by friction will be the difference in the kinetic energy we calculated and (weight * vertical displacement). If we are given the friction force, the work done by this force is f * 'ds. For an object moving down an incline, the work done by the net force is the difference between the work done by gravity and the work done by the friction force. For an object moving up an incline, the work done by the net force is the sum of the work done by the friction force and work done by gravity. confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:45:42 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:56:06 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 --> The ratio of: restoring force of a simple pendulum / weight of the pendulum is approximated displacement from equilibrium / Length of the pendulum Due to the properties of similar triangles. This is only valid for small displacements because for these small displacements the corresponding vertical displacements are negligible and the triangles can be treated as though they are similar. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:56:36 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:56:41 prin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
16:56:45 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 --> self critique assessment:
.................................................
......!!!!!!!!...................................
16:56:49 gen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
16:56:54 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 --> self critique assessment:
.................................................
......!!!!!!!!...................................
17:09:04 **** 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 --> For a man pulling his body weight a vertical displacement of 0.4 m, he exerts 3.92 J/kg: Fnet 'ds = Work = mg 'ds Work/m = g 'ds = 9.8 m/sec^2 * 0.4m = 3.92 m^2 / sec^2 = 3.92 J/kg The working muscles are 5.6% of the man's total body mass: 3.92 / 70 = 0.56 or 5.6 % For a child with arms 1/2 his in length, the pullup only requires force through 0.2 m, and he exerts 1.96 J/kg. He uses half the force per kg, so his working muscles are 2.8% his total body mass. Children do not have to do as much work because they lift a smaller mass through a smaller displacement. confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:09:20 ** 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 --> self critique assessment:
.................................................
......!!!!!!!!...................................
17:38:02 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 --> Since we are given mass, initial velocity, and force, we can use 'dKE = m/2 * (vf^2 - v0^2) to find the final velocity once we calculate the work done by the varying force. The total work done over the displacement will be the integral of the net force function over the interval x = 1 to x = 1.5 The component of the force that contributes to motion in the x-direction is 500*cos (31 degrees) N/m^2 * x^2, which corresponds to 428.6 N/m^2 * x^2. This is the function we'll integrate. The result is 142.87 (1.5^3 -1) = 339.29 Joules. We substitute known values into the equation defining 'dKE to solve for vf = 52.25 m/sec confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:39:43 ** 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 --> This is the answer to #64, which I had trouble with until I realized I was considering the work done by Fx to be *causing* the motion. self critique assessment: 3
.................................................
"