course Phy 121 My email account is not activated. I hope to have this problem resolved by tomorrow evening. Rpײ濨嘿Student Name:
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17:36:44 `q001. Note that there are 4 problems in this set. Two 3 kg masses are suspended over a pulley and a 1 kg mass is added to the mass on one side. Friction exerts a force equal to 2% of the total weight of the system. If the system is given an initial velocity of 5 m/s in the direction of the lighter side, how long will a take the system to come to rest and how far will it travel?
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RESPONSE --> Mass on heavier side = 3 + 1 = 4 kg --> This will be the positive direction of motion. F = 4kg * 9.8m/s = +39.2N Mass on lighter side = 3kg F = 3kg * 9.8m/s = -29.4N --> This force works against the positive direction of motion. fFrict = 2% of total wt Total weight = (4kg + 3kg)(9.8m/s) = 68.6N fFrict = .02(68.6N) = 1.372N Fnet = F + F + fFrict = 39.2 - 29.4 + 1.372 = 11.172 Newtons Acceleration is therefore Fnet/m, a = 11.172N / 7kg = 1.596 m/s^2 The initial velocity is in the direction of the lighter side, which is negative. So initial velocity = -5 m/s. Now we can find change in velocity. `dv = vf - vo = 0m/s -(-5m/s) = 5m/s Now that we know acceleration and change in velocity, we can find the time duration: `dt = `dv/a = 5m/s / 1.596m/s^2 = 3.13 seconds vAve = (vf + vo)/2 = (0 + -5m/s) / 2 = -2.5 m/s. Now we know the time duration and average velocity, we can find the displacement: `ds = vAve * `dt = (-2.5m/s)(3.13s) = -7.825 meters
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17:37:04 We know the initial velocity of the system and, for the period until it comes to rest, we know that its final velocity will be 0 m/s. If we can find the acceleration, we will have three of the five necessary variables with which to analyze the motion and we can therefore determine the time interval `dt and displacement `ds corresponding to this period. We begin by analyzing the forces acting on the system. Before we do so we declare the positive direction of motion for this system to be the direction in which the system moves as the greater of the two hanging masses descends, i.e., the direction of the net force on the system Gravity exerts forces of 4 kg * 9.8 m/s^2 = 39.2 Newtons on the 4 kg mass and 3 kg * 9.8 m/s^2 = 29.4 Newtons on the 3 kg mass. Taking the positive direction to be the direction in which the system moves when the 4 kg mass descends, as stated earlier, then these forces would be +39.2 Newtons and -29.4 Newtons. The total mass of the system is 7 kg, so its total weight is 7 kg * 9.8 m/s^2 = 68.4 Newtons and the frictional force is therefore frictional force = .02 * 68.4 Newtons = 1.4 Newtons, approx.. If the system is moving in the negative direction, then the frictional force is opposed to this direction and therefore positive so the net force on the system is +39.2 Newtons - 29.4 Newtons + 1.4 Newtons = +11.2 Newtons. This results in an acceleration of +11.2 N / (7 kg) = 1.6 m/s^2. We now see that v0 = -5 m/s, vf = 0 and a = 1.6 m/s^2. From this we can easily reason out the desired conclusions. The change in velocity is +5 m/s and the average velocity is -2.5 m/s. At the rate of 1.6 m/s^2, the time interval to change the velocity by +5 m/s is `dt = +5 m/s / (1.6 m/s^2) = 3.1 sec, approx.. At an average velocity of -2.5 m/s, in 3.1 sec the system will be displaced `ds = -2.5 m/s * 3.1 s = -7.8 meters. These conclusions could also have been reached using equations: since vf = v0 + a `dt, `dt - (vf - v0) / a = (0 m/s - (-5 m/s) ) / (1.6 m/s^2) = 3.1 sec (appxox). Since `ds = .5 (v0 + vf) * `dt, `ds = .5 (-5 m/s + 0 m/s) * 3.1 s = -7.8 meters.
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RESPONSE --> I feel pretty good about this one.
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17:38:41 `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds without the application of any external force, then what would be its final displacement relative to its initial position?
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RESPONSE --> If the 5m/s velocity in direction of the three mass and moved for 10 seconds without any other forces, we would use this sequence of operations: The initial velocity will be zero this time. The friction would now oppose the direction of motion, changing one of the first steps: Fnet = 39.2N - 29.4N - 1.372N = 8.428N `dt = 10s - 3.13s = 6.87s a = 8.428N / 7kg = 1.204 m/s^2 We can now find change in velocity: `dv = a * `dt = 1.204m/s^2 * 6.87s = 8.27m/s vf = `dv = 8.27m/s vAve = 8.27m/s / 2 = 4.14m/s Now we can find displacement, since we know `dt and vAve: `ds = (vAve)(`dt) = (4.14m/s)(6.87s) = 28.44 meters
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17:38:52 Since the acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction, if we have both positive and negative velocities this problem must the separated into two parts. As seen in the previous example, in the first 3.2 seconds the system displaces -7.8 meters. This leaves 6.8 seconds after that instant during which the system may accelerate from rest in the positive direction. We therefore analyze the motion from the instant the system comes to rest until the remaining .8 seconds has elapsed. The frictional force during this time will be negative, as it must oppose the direction of motion. The net force on the system will therefore be + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be 8.4 Newtons / (7 kg) = 1.2 m/s^2, approx.. The initial velocity during this phase is 0, the time interval is 6.8 sec and the acceleration is 1.2 m/s^2. We therefore conclude that the velocity will change by 1.2 m/s^2 * 6.8 sec = 8.2 m/s, approx, ending up at 8.2 m/s since this phase started at 0 m/s. This gives an average velocity of 4.1 m/s; during 6.8 sec the object therefore displaces `ds = 4.1 m/s * 6.8 sec = 28 meters approx.. These results could have also been easily obtained from equations. For the entire 10 seconds, the displacements were -7.8 meters and +28 meters, for a net displacement of approximately +20 meters. That is, the system is at this instant about 20 meters in the direction of the 4 kg mass from its initial position.
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RESPONSE --> I understand this problem.
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17:44:04 `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)?
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RESPONSE --> vo = 5m/s `ds = 100 m Mass = 1400kg Weight = 1400kg * 9.8m/s^2 = 13720 N fFrict = 0.1 x normal force = 0.1 * 13720N = 137.2N = -137.2, against direction of motion Incline slope = 0.05 or 5% parallel wt = 13720N * .05 = 686 N Fnet = 686N - 137.2 = 548.8N We can find the final velocity by finding the acceleration first: acceleration = Fnet/m = 548.8N/1400kg = 0.392 m/s^2 Now we can say that vf^2 = vo^2 + 2a*`ds vf^2 = (5m/s)^2 + 2(0.392m/s^2)(100m) = 103.4 m^2/s^2 TAKE SQUARE ROOT OF BOTH SIDES: vf = +- 10.17m/s, will be positive because both initial velocity and acceleration are positive.
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17:44:20 We are given the initial velocity and displacement of the automobile, so we need only find the acceleration and we can analyzes problem as a standard uniform acceleration problem. The automobile experiences a net force equal to the component of its weight which is parallel to the incline plus the force of friction. If we regard the direction down the incline as positive, the parallel component of the weight will be positive and the frictional force, which must be in the direction opposite that of the velocity, will be negative. The weight of the automobile is 1400 kg * 9.8 m/s^2 = 13720 Newtons, so the component of the weight parallel to the incline is parallel weight component = 13720 Newtons * .05 = 686 Newtons. The normal force between the road and the tires is very nearly equal to the weight of the car because of the small slope, so the magnitude of the frictional force is approximately frictional force = 13720 Newtons * .01 = 137 Newtons, approx.. The frictional force is therefore -137 Newtons and the net force on the automobile is Fnet = 686 Newtons - 137 Newtons = 550 Newtons (approx.). It follows that the acceleration of the automobile must be a = Fnet / m = 550 Newtons / 1400 kg = .4 m/s^2 (approx.). We now have a uniform acceleration problem with initial velocity v0 = 5 meters/second, displacement `ds = 100 meters and acceleration a = .4 m/s^2. We can easily find the final velocity using the equation vf^2 = v0^2 + 2 a `ds, which gives us vf = +- `sqrt(v0^2 + 2 a `ds) = +- `sqrt( (5 m/s)^2 + 2 * .4 m/s^2 * 100 m) = +- `sqrt( 25 m^2 / s^2 + 80 m^2 / s^2) = +-`sqrt(105 m^2 / s^2) = +- 10.2 m/s. It is obvious that the final velocity in this problem is the positive solution +10.2 m/s, since the initial velocity and acceleration are both positive.
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RESPONSE --> Got it!
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17:45:35 `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast?
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RESPONSE --> vo = 11.2m/s I will assume for this problem, if the automobile is going uphill is moving in the positive direction which is opposite its normal forces, such as gravity. Therefore the net force will most likely be negative. Nothing else has changed, so we need to correct the direction of these forces and work with the new initial velocity. Fnet = -686N - 137.2N = -823.2 Newtons Now acceleration = -823.2N/1400kg = -0.588 m/s^2 Change in velocity = vf - vo = 0m/s - 11.2m/s = -11.2 m/s Now the time duration = `dv/a = (-11.2m/s)/(-0.588m/s^2) = 19.05 seconds. Average velocity = (11.2m/s)/2 = 5.6m/s Displacement = vAve * `dt = (5.6m/s)(19.05s) = 106.68 meters
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17:45:55 We again have a uniform acceleration situation where the initial velocity is known and we have the information to determine the acceleration. The other quantity we can deduce is the final velocity, which at the furthest point up the hill will be zero. Since the automobile is coasting up the incline, we will take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, so will therefore be negative. The parallel component of the weight will be the same as before, 686 Newtons, but being directed down the incline will be in the direction opposite to that if the velocity and will therefore also be negative. The net force on the automobile therefore be net force = -686 Newtons - 137 Newtons = -820 Newtons (approx.). Its acceleration will be a = Fnet / m = -820 Newtons / 1400 kg = -.6 m/s^2 (approx.). We see now that v0 = 11.2 m/s, vf = 0 and a = -.6 m/s^2. Either by direct reasoning or by using an equation we easily find that `dt = (-11.2 m/s) / (-.6 m/s^2) = 19 sec (approx) and `ds = (11.2 m/s + 0 m/s) / 2 * 19 sec = 5.6 m/s * 19 sec = 106 meters (approx).
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RESPONSE --> Ok
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ԀJ㽕yާr assignment #012 012. `query 12 Physics I 06-21-2006
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21:48:19 Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?
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RESPONSE --> Gravitational forces act on the system, but not frictional since there is no friction. The net force of the system is the product of the hanging mass, m2, and the gravitational force, 9.8m/s^2. The acceleration of the system would be equal to the net force divided by the total mass. (Fnet / (m1 + m2) ). If the hanging mass descended a distance `dy, the change in PE would be the negative product of work by gravity on the system. `dPE = - (9.8m/s^2 * m2)(`dy).
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21:49:03 ** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet. Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2). If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy. COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS: The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension. INSTRUCTOR COMMENT: String tension shouldn't be counted among the forces contributing to the net force on the system. The string tension is internal to the two-mass system. It doesn't act on the system but within the system. Net force is therefore suspended mass * accel due to gravity only 'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **
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RESPONSE --> Ok
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21:49:27 How would friction change your answers to the preceding question?
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RESPONSE --> Friction would be working in the opposite direction of m1 and its gravitational force, if any. Fnet = F of m2 - fFriction
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21:49:37 **Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **
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RESPONSE --> I understand this problem.
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21:50:24 Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.
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RESPONSE --> Potential energy can be found by finding the area under the graph of force vs stretch.
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21:51:00 ** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches. So the PE stored is the area under the graph of force vs. stretch. **
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RESPONSE --> Ok, I have a little trouble understanding the details of this problem. NOTE: This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position.
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21:51:52 STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?
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RESPONSE --> The area under the graph of force vs. stretch represents the potential energy at given stretch, it is therefore, also equivalent to work done by or on the system depending on the situation.
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21:52:52 ** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement). If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it. If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **
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RESPONSE --> This description helps my understanding.
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21:53:45 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I had a hard time at first, seeing how the potential energy, 3.3962 Joules, was found in the introductory problem #14.
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course Phy 121 My email account is not activated. I hope to have this problem resolved by tomorrow evening. Rpײ濨嘿Student Name:
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17:36:44 `q001. Note that there are 4 problems in this set. Two 3 kg masses are suspended over a pulley and a 1 kg mass is added to the mass on one side. Friction exerts a force equal to 2% of the total weight of the system. If the system is given an initial velocity of 5 m/s in the direction of the lighter side, how long will a take the system to come to rest and how far will it travel?
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RESPONSE --> Mass on heavier side = 3 + 1 = 4 kg --> This will be the positive direction of motion. F = 4kg * 9.8m/s = +39.2N Mass on lighter side = 3kg F = 3kg * 9.8m/s = -29.4N --> This force works against the positive direction of motion. fFrict = 2% of total wt Total weight = (4kg + 3kg)(9.8m/s) = 68.6N fFrict = .02(68.6N) = 1.372N Fnet = F + F + fFrict = 39.2 - 29.4 + 1.372 = 11.172 Newtons Acceleration is therefore Fnet/m, a = 11.172N / 7kg = 1.596 m/s^2 The initial velocity is in the direction of the lighter side, which is negative. So initial velocity = -5 m/s. Now we can find change in velocity. `dv = vf - vo = 0m/s -(-5m/s) = 5m/s Now that we know acceleration and change in velocity, we can find the time duration: `dt = `dv/a = 5m/s / 1.596m/s^2 = 3.13 seconds vAve = (vf + vo)/2 = (0 + -5m/s) / 2 = -2.5 m/s. Now we know the time duration and average velocity, we can find the displacement: `ds = vAve * `dt = (-2.5m/s)(3.13s) = -7.825 meters
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17:37:04 We know the initial velocity of the system and, for the period until it comes to rest, we know that its final velocity will be 0 m/s. If we can find the acceleration, we will have three of the five necessary variables with which to analyze the motion and we can therefore determine the time interval `dt and displacement `ds corresponding to this period. We begin by analyzing the forces acting on the system. Before we do so we declare the positive direction of motion for this system to be the direction in which the system moves as the greater of the two hanging masses descends, i.e., the direction of the net force on the system Gravity exerts forces of 4 kg * 9.8 m/s^2 = 39.2 Newtons on the 4 kg mass and 3 kg * 9.8 m/s^2 = 29.4 Newtons on the 3 kg mass. Taking the positive direction to be the direction in which the system moves when the 4 kg mass descends, as stated earlier, then these forces would be +39.2 Newtons and -29.4 Newtons. The total mass of the system is 7 kg, so its total weight is 7 kg * 9.8 m/s^2 = 68.4 Newtons and the frictional force is therefore frictional force = .02 * 68.4 Newtons = 1.4 Newtons, approx.. If the system is moving in the negative direction, then the frictional force is opposed to this direction and therefore positive so the net force on the system is +39.2 Newtons - 29.4 Newtons + 1.4 Newtons = +11.2 Newtons. This results in an acceleration of +11.2 N / (7 kg) = 1.6 m/s^2. We now see that v0 = -5 m/s, vf = 0 and a = 1.6 m/s^2. From this we can easily reason out the desired conclusions. The change in velocity is +5 m/s and the average velocity is -2.5 m/s. At the rate of 1.6 m/s^2, the time interval to change the velocity by +5 m/s is `dt = +5 m/s / (1.6 m/s^2) = 3.1 sec, approx.. At an average velocity of -2.5 m/s, in 3.1 sec the system will be displaced `ds = -2.5 m/s * 3.1 s = -7.8 meters. These conclusions could also have been reached using equations: since vf = v0 + a `dt, `dt - (vf - v0) / a = (0 m/s - (-5 m/s) ) / (1.6 m/s^2) = 3.1 sec (appxox). Since `ds = .5 (v0 + vf) * `dt, `ds = .5 (-5 m/s + 0 m/s) * 3.1 s = -7.8 meters.
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RESPONSE --> I feel pretty good about this one.
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17:38:41 `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds without the application of any external force, then what would be its final displacement relative to its initial position?
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RESPONSE --> If the 5m/s velocity in direction of the three mass and moved for 10 seconds without any other forces, we would use this sequence of operations: The initial velocity will be zero this time. The friction would now oppose the direction of motion, changing one of the first steps: Fnet = 39.2N - 29.4N - 1.372N = 8.428N `dt = 10s - 3.13s = 6.87s a = 8.428N / 7kg = 1.204 m/s^2 We can now find change in velocity: `dv = a * `dt = 1.204m/s^2 * 6.87s = 8.27m/s vf = `dv = 8.27m/s vAve = 8.27m/s / 2 = 4.14m/s Now we can find displacement, since we know `dt and vAve: `ds = (vAve)(`dt) = (4.14m/s)(6.87s) = 28.44 meters
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17:38:52 Since the acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction, if we have both positive and negative velocities this problem must the separated into two parts. As seen in the previous example, in the first 3.2 seconds the system displaces -7.8 meters. This leaves 6.8 seconds after that instant during which the system may accelerate from rest in the positive direction. We therefore analyze the motion from the instant the system comes to rest until the remaining .8 seconds has elapsed. The frictional force during this time will be negative, as it must oppose the direction of motion. The net force on the system will therefore be + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be 8.4 Newtons / (7 kg) = 1.2 m/s^2, approx.. The initial velocity during this phase is 0, the time interval is 6.8 sec and the acceleration is 1.2 m/s^2. We therefore conclude that the velocity will change by 1.2 m/s^2 * 6.8 sec = 8.2 m/s, approx, ending up at 8.2 m/s since this phase started at 0 m/s. This gives an average velocity of 4.1 m/s; during 6.8 sec the object therefore displaces `ds = 4.1 m/s * 6.8 sec = 28 meters approx.. These results could have also been easily obtained from equations. For the entire 10 seconds, the displacements were -7.8 meters and +28 meters, for a net displacement of approximately +20 meters. That is, the system is at this instant about 20 meters in the direction of the 4 kg mass from its initial position.
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RESPONSE --> I understand this problem.
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17:44:04 `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)?
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RESPONSE --> vo = 5m/s `ds = 100 m Mass = 1400kg Weight = 1400kg * 9.8m/s^2 = 13720 N fFrict = 0.1 x normal force = 0.1 * 13720N = 137.2N = -137.2, against direction of motion Incline slope = 0.05 or 5% parallel wt = 13720N * .05 = 686 N Fnet = 686N - 137.2 = 548.8N We can find the final velocity by finding the acceleration first: acceleration = Fnet/m = 548.8N/1400kg = 0.392 m/s^2 Now we can say that vf^2 = vo^2 + 2a*`ds vf^2 = (5m/s)^2 + 2(0.392m/s^2)(100m) = 103.4 m^2/s^2 TAKE SQUARE ROOT OF BOTH SIDES: vf = +- 10.17m/s, will be positive because both initial velocity and acceleration are positive.
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17:44:20 We are given the initial velocity and displacement of the automobile, so we need only find the acceleration and we can analyzes problem as a standard uniform acceleration problem. The automobile experiences a net force equal to the component of its weight which is parallel to the incline plus the force of friction. If we regard the direction down the incline as positive, the parallel component of the weight will be positive and the frictional force, which must be in the direction opposite that of the velocity, will be negative. The weight of the automobile is 1400 kg * 9.8 m/s^2 = 13720 Newtons, so the component of the weight parallel to the incline is parallel weight component = 13720 Newtons * .05 = 686 Newtons. The normal force between the road and the tires is very nearly equal to the weight of the car because of the small slope, so the magnitude of the frictional force is approximately frictional force = 13720 Newtons * .01 = 137 Newtons, approx.. The frictional force is therefore -137 Newtons and the net force on the automobile is Fnet = 686 Newtons - 137 Newtons = 550 Newtons (approx.). It follows that the acceleration of the automobile must be a = Fnet / m = 550 Newtons / 1400 kg = .4 m/s^2 (approx.). We now have a uniform acceleration problem with initial velocity v0 = 5 meters/second, displacement `ds = 100 meters and acceleration a = .4 m/s^2. We can easily find the final velocity using the equation vf^2 = v0^2 + 2 a `ds, which gives us vf = +- `sqrt(v0^2 + 2 a `ds) = +- `sqrt( (5 m/s)^2 + 2 * .4 m/s^2 * 100 m) = +- `sqrt( 25 m^2 / s^2 + 80 m^2 / s^2) = +-`sqrt(105 m^2 / s^2) = +- 10.2 m/s. It is obvious that the final velocity in this problem is the positive solution +10.2 m/s, since the initial velocity and acceleration are both positive.
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RESPONSE --> Got it!
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17:45:35 `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast?
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RESPONSE --> vo = 11.2m/s I will assume for this problem, if the automobile is going uphill is moving in the positive direction which is opposite its normal forces, such as gravity. Therefore the net force will most likely be negative. Nothing else has changed, so we need to correct the direction of these forces and work with the new initial velocity. Fnet = -686N - 137.2N = -823.2 Newtons Now acceleration = -823.2N/1400kg = -0.588 m/s^2 Change in velocity = vf - vo = 0m/s - 11.2m/s = -11.2 m/s Now the time duration = `dv/a = (-11.2m/s)/(-0.588m/s^2) = 19.05 seconds. Average velocity = (11.2m/s)/2 = 5.6m/s Displacement = vAve * `dt = (5.6m/s)(19.05s) = 106.68 meters
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17:45:55 We again have a uniform acceleration situation where the initial velocity is known and we have the information to determine the acceleration. The other quantity we can deduce is the final velocity, which at the furthest point up the hill will be zero. Since the automobile is coasting up the incline, we will take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, so will therefore be negative. The parallel component of the weight will be the same as before, 686 Newtons, but being directed down the incline will be in the direction opposite to that if the velocity and will therefore also be negative. The net force on the automobile therefore be net force = -686 Newtons - 137 Newtons = -820 Newtons (approx.). Its acceleration will be a = Fnet / m = -820 Newtons / 1400 kg = -.6 m/s^2 (approx.). We see now that v0 = 11.2 m/s, vf = 0 and a = -.6 m/s^2. Either by direct reasoning or by using an equation we easily find that `dt = (-11.2 m/s) / (-.6 m/s^2) = 19 sec (approx) and `ds = (11.2 m/s + 0 m/s) / 2 * 19 sec = 5.6 m/s * 19 sec = 106 meters (approx).
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RESPONSE --> Ok
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ԀJ㽕yާr assignment #012 012. `query 12 Physics I 06-21-2006
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21:48:19 Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?
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RESPONSE --> Gravitational forces act on the system, but not frictional since there is no friction. The net force of the system is the product of the hanging mass, m2, and the gravitational force, 9.8m/s^2. The acceleration of the system would be equal to the net force divided by the total mass. (Fnet / (m1 + m2) ). If the hanging mass descended a distance `dy, the change in PE would be the negative product of work by gravity on the system. `dPE = - (9.8m/s^2 * m2)(`dy).
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21:49:03 ** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet. Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2). If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy. COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS: The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension. INSTRUCTOR COMMENT: String tension shouldn't be counted among the forces contributing to the net force on the system. The string tension is internal to the two-mass system. It doesn't act on the system but within the system. Net force is therefore suspended mass * accel due to gravity only 'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **
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RESPONSE --> Ok
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21:49:27 How would friction change your answers to the preceding question?
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RESPONSE --> Friction would be working in the opposite direction of m1 and its gravitational force, if any. Fnet = F of m2 - fFriction
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21:49:37 **Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **
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RESPONSE --> I understand this problem.
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21:50:24 Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.
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RESPONSE --> Potential energy can be found by finding the area under the graph of force vs stretch.
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21:51:00 ** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches. So the PE stored is the area under the graph of force vs. stretch. **
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RESPONSE --> Ok, I have a little trouble understanding the details of this problem. NOTE: This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position.
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21:51:52 STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?
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RESPONSE --> The area under the graph of force vs. stretch represents the potential energy at given stretch, it is therefore, also equivalent to work done by or on the system depending on the situation.
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21:52:52 ** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement). If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it. If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **
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RESPONSE --> This description helps my understanding.
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21:53:45 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I had a hard time at first, seeing how the potential energy, 3.3962 Joules, was found in the introductory problem #14.
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