course phy201 assignment #016016. `query 16
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19:28:41 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
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RESPONSE --> so much of the kinetic enegy was used to create the displacement confidence assessment: 1
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19:28:58 ** This question should have specified just the KE in the vertical direction. The kinetic energy of the ball in the vertical direction will be proportional to `dy. The reason: The vertical velocity attained by the ball is vf = `sqrt(v0^2 + 2 a `ds). Since the initial vertical velocity is 0, for distance of fall `dy we have vf = `sqrt( 2 a `dy ), showing that the vertical velocity is proportional to the square root of the distance fallen. Since KE is .5 m v^2, the KE will be proportional to the square of the velocity, hence to the square of the square root of `dy. Thus KE is proportional to `dy. **
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RESPONSE --> self critique assessment: 1
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19:30:25 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
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RESPONSE --> kinetic enegy is calculated actual energy while PE is the potiental energy that an object has. confidence assessment: 0
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19:30:45 ** STUDENT RESPONSE: Because actually some of the energy will be dissapated in the rotation of the ball as it drops? INSTRUCTOR COMMENT: Good, but note that rotation doesn't dissipate KE, it merely accounts for some of the KE. Rotational KE is recoverable--for example if you place a spinning ball on an incline the spin can carry the ball a ways up the incline, doing work in the process. The PE loss is converted to KE, some into rotational KE which doesn't contribute to the range of the ball and some of which simply makes the ball spin. ANOTHER STUDENT RESPONSE: And also the loss of energy due to friction and conversion to thermal energy. INSTRUCTOR COMMENT: Good. There would be a slight amount of air friction and this would dissipate energy as you describe here, as would friction with the ramp (which would indeed result in dissipation in the form of thermal energy). **
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RESPONSE --> self critique assessment: 1
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19:32:23 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> 132 joules confidence assessment: 0
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19:33:03 The work required to stop the automobile, by the work-energy theorem, is equal and opposite to its change in kinetic energy: `dW = - `dKE. The initial KE of the automobile is .5 m v^2, and before calculating this we convert 105 km/hr to m/s: 105 km/hr = 105 km / hr * 1000 m / km * 1 hr / 3600 s = 29.1 m/s. Our initial KE is therefore KE = .5 m v^2 = .5 * 1250 kg * (29.1 m/s)^2 = 530,000 kg m^2 / s^2 = 530,000 J. The car comes to rest so its final KE is 0. The change in KE is therefore -530,000 J. It follows that the work required to stop the car is `dW = - `dKE = - (-530,000 J) = 530,000 J.
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RESPONSE --> the correct answer is 530,000 joules self critique assessment: 1
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19:33:28 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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RESPONSE --> confidence assessment: 0
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19:33:51 The force exerted by a spring at equilibrium is 0, and the force at position x is - k x, so the average force exerted between equilibrium and position x is (0 + (-kx) ) / 2 = -1/2 k x. The work done by the spring as it is stretched from equilibrium to position x, a displacment of x, is therefore `dW = F * `ds = -1/2 k x * x = -1/2 k x^2. The only force exerted by the spring is the conservative elastic force, so the PE change of the spring is therefore `dPE = -`dW = - (-1/2 kx^2) = 1/2 k x^2. That is, the spring stores PE = 1/2 k x^2. In this situation k = 440 N / m and the desired PE is 25 J. Solving PE = 1/2 k x^2 for x (multiply both sides by 2 and divide both sides by k, then take the square root of both sides) we obtain x = +-sqrt(2 PE / k) = +-sqrt( 2 * 25 J / (440 N/m) ) = +- sqrt( 50 kg m^2 / s^2 / ( (440 kg m/s^2) / m) )= +- sqrt(50 / 440) sqrt(kg m^2 / s^2 * (s^2 / kg) ) = +- .34 sqrt(m^2) = +-.34 m. The spring will store 25 J of energy at either the +.34 m or the -.34 m position.
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RESPONSE --> self critique assessment:
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19:36:09 gen phy text problem 6.19 88 g arrow 78 cm ave force 110 N, speed? What did you get for the speed of the arrow?
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RESPONSE --> 97.5cm/s confidence assessment: 1
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19:36:41 ** 110 N acting through 78 cm = .78 m does work `dW = 110 N * .78 m = 86 Joules appxo.. If all this energy goes into the KE of the arrow then we have a mass of .088 kg with 86 Joules of KE. We can solve .5 m v^2 = KE for v, obtaining | v | = sqrt( 2 * KE / m) = sqrt(2 * 86 Joules / (.088 kg) ) = sqrt( 2000 kg m^2 / s^2 * 1 / kg) = sqrt(2000 m^2 / s^2) = 44 m/s, approx.. **
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RESPONSE --> self critique assessment: 0
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19:37:10 query univ phy 6.84 (6.74 10th edition) bow full draw .75 m, force from 0 to 200 N to 70 N approx., mostly concave down. What will be the speed of the .0250 kg arrow as it leaves the bow?
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RESPONSE --> confidence assessment: 1
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19:38:08 ** The work done to pull the bow back is represented by the area beneath the force vs. displacement curve. The curve could be approximated by a piecewise straight line from about 0 to 200 N then back to 70 N. The area beneath this graph would be about 90 N m or 90 Joules. The curve itself probably encloses a bit more area than the straight line, so let's estimate 100 Joules (that's probably a little high, but it's a nice round number). If all the energy put into the pullback goes into the arrow then we have a .0250 kg mass with kinetic energy 100 Joules. Solving KE = .5 m v^2 for v we get v = sqrt(2 KE / m) = sqrt( 2 * 100 Joules / ( .025 kg) ) = sqrt(8000 m^2 / s^2) = 280 m/s, approx. **
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RESPONSE --> self critique assessment: 1
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18:55:25 `q001. An object of mass 10 kg is subjected to a net force of 40 Newtons as it accelerates from rest through a distance of 20 meters. Find the final velocity of the object.
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RESPONSE --> m = 10 kg, Fnet = 40 N, 'ds = 20 m, v0 = 0 Find vf, F = m * a a = 40/10 = 4m/s^2, vf^2 = v0^2 + (2 * a * 'ds) vf^2 = 0 +(2 * 4 * 20) = 160 vf^2 = 160 vf = 12.65 m/s confidence assessment: 3
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18:55:57 We know the initial velocity v0 = 0 and the displacement `ds = 20 meters. We have the information we need to determine the acceleration of the object. Once we find that acceleration we can easily determine its final velocity vf. We first find the acceleration. The object is subjected to a net force of 40 Newtons and has mass 10 kg, so that will have acceleration a = Fnet / m = 40 Newtons / 10 kg = 4 m/s^2. We can use the equation vf^2 = v0^2 + 2 a `ds to see that vf = +- `sqrt( v0^2 + 2 a `ds ) = +- `sqrt ( 0 + 2 * 4 m/s^2 * 20 meters) = +-`sqrt(160 m^2 / s^2) = +-12.7 m/s. The acceleration and displacement have been taken to be positive, so the final velocity will also be positive and we see that vf = + 12.7 m/s.
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RESPONSE --> Looks good to me self critique assessment: 3
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19:04:13 `q002. Find the value of the quantity 1/2 m v^2 at the beginning of the 20 meter displacement, the value of the same quantity at the end of this displacement, and the change in the quantity 1/2 m v^2 for this displacement.
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RESPONSE --> 'ds = 20 m, v0 = 0, vf = 12.65 m/s, m = 10 kg Find KE = 1/2 * m * v^2 KEv0 = 0 N KE = 1/2 * 10 * 12.65^2 = 5 * 160 = 800 KE = 800 Joules confidence assessment: 3
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19:05:14 Over the 20 meter displacement the velocity changes from v0 = 0 m/s to vf = 12.7 m/s. Thus the quantity 1/2 m v^2 changes from initial value 1/2 (10 kg) (0 m/s)^2 = 0 to final value 1/2 (10 kg)(12.7 m/s)^2 = 800 kg m^2 / s^2.
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RESPONSE --> Still OK for me self critique assessment: 3
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19:10:07 `q003. Find the value of the quantity Fnet * `ds for the present example, and express this quantity in units of kg, meters and seconds.
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RESPONSE --> Fnet * 'ds in quanity terms of kg, m, s Fnet = m *a, so m * a * 'ds = 10 kg * 4 m/s^2 * 20 m = 800 (kg m^2)/s^2 confidence assessment: 3
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19:10:49 Fnet = 40 Newtons and `ds = 20 meters, so Fnet * `ds = 40 Newtons * 20 meters = 800 Newton meters. Recall that a Newton (being obtained by multiplying mass in kg by acceleration in m/s^2) is a kg * m/s^2, so that the 800 Newton meters can be expressed as 800 kg m/s^2 * meters = 800 kg m^2 / s^2.
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RESPONSE --> Still feeling pretty good self critique assessment: 3
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19:18:44 `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
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RESPONSE --> Actually the quantities get the same answer when there isn't any friction or opposite forces to deal with because Fnet * 'ds = 'dW and 1/2 * m v^2 = KE so ;dW = KE in this instance. confidence assessment: 1
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19:20:11 The change in the quantity Fnet * `ds is 800 kg m^2 / s^2 and the change in 1/2 m v^2 is 800 kg m^2 / s^2. The quantities are therefore the same. This quantity could also be expressed as 800 Newton meters, as it was in the initial calculation of the less question. We define 1 Joule to be 1 Newton * meter, so that the quantity 800 Newton meters is equal to 800 kg m^2 / s^2 and also equal to 800 Joules.
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RESPONSE --> I suppose my thinking was the same without the 800 Joules explanation self critique assessment: 2
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19:57:05 `q005. Suppose that all the quantities given in the previous problem are the same except that the initial velocity is 9 meters / second. Again calculate the final velocity, the change in (1/2 m v^2) and Fnet * `ds.
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RESPONSE --> v0=9m/s, m=10kg, Fnet=40 N, a=4m/s^2 'ds=20m, vf^2=v0^2 + (2*a*'ds) vf^2 = (9)^2 + (2 * 4 * 20) vf^2 = 81 + 160 = 241 vf^2 = 241 vf = 15.52 m/s 'dW = Fnet * 'ds, 'dW = 40 N * 20 m = 800J KE = 1/2 * m * vf^2 - 1/2 * m * v0^2 KE = 1/2*10 kg*(15.52)^2-1/2*10kg*(9)^2 KE = 1204 - 405 = 799 Joules *** So 'dW is still equal to the 'dKE confidence assessment: 1
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20:01:30 The acceleration results from the same net force acting on the same mass so is still 4 m/s^2. This time the initial velocity is v0 =9 m/s, and the displacement is still `ds = 20 meters. We therefore obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt( (9 m/s)^2 + 2 * 4 m/s^2 * 20 meters) = +_`sqrt( 81 m^2 / s^2 + 160 m^2 / s^2) = +_`sqrt( 241 m^2 / s^2) = +_15.5 m/s (approx). For the same reasons as before we choose the positive velocity +15.5 m/s. The quantity 1/2 mv^2 is initially 1/2 * 10 kg * (9 m/s)^2 = 420 kg m^2 / s^2 = 420 Joules, and reaches a final value of 1/2 * 10 kg * (15.5 m/s)^2 = 1220 kg m^2 /s^2 = 1220 Joules (note that this value is obtained using the accurate value `sqrt(241) m/s rather than the approximate 15.5 m/s; if the rounded-off approximation 15.5 m/s is used, the result will differ slightly from 1220 Joules). The quantity therefore changes from 420 Joules to 1220 Joules, a change of +800 Joules. The quantity Fnet * `ds is the same as in the previous exercise, since Fnet is still 40 Newtons and `ds is still 20 meters. Thus Fnet * `ds = 800 Joules. We see that, at least for this example, the change in the quantity 1/2 m v^2 is equal to the product Fnet * `ds. We ask in the next problem if this will always be the case for any Fnet, mass m and displacement `ds. [Important note: When we find the change in the quantity 1/2 m v^2 we calculate 1/2 m v^2 for the initial velocity and then again for the final velocity and subtract in the obvious way. We do not find a change in the velocity and plug that change into 1/2 m v^2. If we had done so with this example we would have obtained about 205 Joules, much less than the 800 Joules we obtain if we correctly find the difference in 1/2 m v^2. Keep this in mind. The quantity 1/2 m v^2 is never calculated using a difference in velocities for v; it works only for actual velocities.]
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RESPONSE --> I definitely did the calculation both ways in finding the initial KE and the final KE. I was still looking for the net 800J from the previoius problem which indicated to me that you have to calculate the KEv0 and KE vf separately self critique assessment: 2
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20:22:17 `q006. The quantity Fnet * `ds and the change in the quantity 1/2 m v^2 were the same in the preceding example. This might be just a coincidence of the numbers chosen, but if so we probably wouldn't be making is bigger deal about it. In any case if the numbers were just chosen at random and we obtained this sort of equality, we would be tempted to conjecture that the quantities were indeed always equal. Answer the following: How could we determine if this conjecture is correct? Hint: Let Fnet, m and `ds stand for any net force, mass and displacement and let v0 stand for any initial velocity. In terms of these symbols obtain the expression for v0 and vf, then obtain the expression for the change in the quantity1/2 m v^2. See if the result is equal to Fnet * `ds.
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RESPONSE --> F = m * a, a = F/m, vf^2 = v0^2 + 2 * a * 'ds vf^2 = v0^2 + (2 * F/m * 'ds)
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20:27:08 Following the same order of reasoning as used earlier, we see that the expression for the acceleration is a = Fnet / m. If we assume that v0 and `ds are known then once we have acceleration a we can use vf^2 = v0^2 + 2 a `ds to find vf. This is good because we want to find an expression for 1/2 m v0^2 and another for 1/2 m vf^2. First we substitute Fnet / m for a and we obtain vf^2 = v0^2 + 2 * Fnet / m * `ds. We can now determine the values of 1/2 m v^2 for v=v0 and v=vf. For v = v0 we obtain 1/2 m v0^2; this expression is expressed in terms of the four given quantities Fnet, m, `ds and v0, so we require no further change in this expression. For v = vf we see that 1/2 m v^2 = 1/2 m vf^2. However, vf is not one of the four given symbols, so we must express this as 1/2 m vf^2 = 1/2 m (v0^2 + 2 Fnet/m * `ds). Now the change in the quantity 1/2 m v^2 is change in 1/2 m v^2: 1/2 m vf^2 - 1/2 m v0^2 = 1/2 m (v0^2 + 2 Fnet / m * `ds) - 1/2 m v0^2. Using the distributive law of multiplication over addition we see that this expression is the same as change in 1/2 mv^2: 1/2 m v0^2 + 1/2 * m * 2 Fnet / m * `ds - 1/2 m v0^2, which can be rearranged to 1/2 m v0^2 - 1/2 m v0^2 + 1/2 * m * 2 Fnet / m * `ds = 1/2 * 2 * m * Fnet / m * `ds = Fnet * `ds. Thus we see that for any Fnet, m, v0 and `ds, the change in 1/2 m v^2 must be equal to Fnet * `ds.
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RESPONSE --> I think I see the substitution into the equations but it's knowing what to solve for the find the equality. I will probably know more as time passes on being able to determine this path. self critique assessment: 1
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20:35:17 `q007. We call the quantity 1/2 m v^2 the Kinetic Energy, often abbreviated KE, of the object. We call the quantity Fnet * `ds the work done by the net force, often abbreviated here as `dWnet. Show that for a net force of 12 Newtons and a mass of 48 kg, the work done by the net force in accelerating an object from rest through a displacement of 100 meters is equal to the change in the KE of the mass.
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RESPONSE --> KE = 1/2 * m * v^2, 'dW = Fnet * 'ds Fnet = 12 N, m = 48 kg, v0 = 0, 'ds = 100 m 'dw = 12 N * 100 m = 1200 Joules F = m * a, a = 12 N /48 kg = .25 m/s^2 vf^2 = v0^2 + (2 * a * 'ds) vf^2 = 0 + (2 * .25 * 100) = 50 vf = 7.07 m/s KE = 1/2 * 48 * 7.07^2 = 1199.6 Joules 1199.6 Joules of KE is close enough to the 1200 Joules of 'dW confidence assessment: 2
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20:36:48 The work done by a 12 Newton force acting through a displacement of 100 meters is 12 Newtons * 100 meters = 1200 Newton meters = 1200 Joules. A 48 kg object subjected to a net force of 12 Newtons will accelerate at the rate a = Fnet / m = 12 Newtons / 48 kg = .25 m/s^2. Starting from rest and accelerating through a displacement of 100 meters, this object attains final velocity vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt( 0^2 + 2 * .25 m/s^2 * 100 m) = +-`sqrt(50 m^2/s^2) = 7.1 m/s (approx.). Its KE therefore goes from KE0 = 1/2 m v0^2 = 0 to KEf = 1/2 m vf^2 = 1/2 (48 kg) (7.1 m/s)^2 = 1200 kg m^2/s^2 = 1200 Joules. This is the same quantity calculated usin Fnet * `ds.Thus the change in kinetic energy is equal to the work done.
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RESPONSE --> I suppose I need to say the change in KE is equal to the work self critique assessment: 2
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20:44:02 `q008. How much work is done by the net force when an object of mass 200 kg is accelerated from 5 m/s to 10 m/s? Find your answer without using the equations of motion.
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RESPONSE --> m =200 kg, v0 = 5 m/s, vf = 10 m/s KEvf = 1/2 * 200 * 10^2 = 10000 Joules KEv0 = 1/2 * 200 * 5^2 = 2500 Joules KEnet = 10000 - 2500 = 7500 Joules KEnet = 'dW = 7500 Joules of work done confidence assessment: 2
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20:45:49 The work done by the net force is equal to the change in the KE of the object. The initial kinetic energy of the object is KE0 = 1/2 m v0^2 = 1/2 (200 kg) (5 m/s)^2 = 2500 kg m^2/s^2 = 2500 Joules. The final kinetic energy is KEf = 1/2 m vf^2 = 1/2 (200 kg)(10 m/s)^2 = 10,000 Joules. The change in the kinetic energy is therefore 10,000 Joules - 2500 Joules = 7500 Joules. The same answer would have been calculated calculating the acceleration of the object, which because of the constant mass and constant net force is uniform, the by using the equations of motion to determine the displacement of the object, the multiplying by the net force.
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RESPONSE --> I had to think if the KE formula was a motion formula but since it had to do with work and energy it was a motion formula self critique assessment: 2
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20:52:41 `q009. Answer the following without using the equations of uniformly accelerated motion: If the 200 kg object in the preceding problem is uniformly accelerated from 5 m/s to 10 m/s while traveling 50 meters, then what net force was acting on the object?
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RESPONSE --> m = 200kg, v0 = 5 m/s, vf = 10 m/s, 'ds = 50 m Find force without motion formulas KE = 1/2 * m * v^2, 'dW = Fnet * 'ds KEvf = 1/2 * 200 * 10^2 = 10000 Joules KEv0 = 1/2 * 200 * 5^2 = 2500 Joules KEnet = 10000 - 2500 = 7500 Joules KEnet = 'dW = 7500 Joules of work done 'dW = Fnet * 'ds, Fnet = 'dW/'ds Fnet = 7500/50 = 150 Newtons Fnet = 150 Newtons confidence assessment: 2
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20:54:20 The net force did 7500 Joules of work. Since the object didn't change mass and since its acceleration was constant, the net force must have been constant. So the work done was `dWnet = Fnet * `ds = 7500 Joules. Since we know that `ds is 50 meters, we can easily solve for Fnet: Fnet = `dWnet / `ds = 7500 Joules / 50 meters = 150 Newtons. [Note that this problem could have been solved using the equations of motion to find the acceleration of the object, which could then have been multiplied by the mass of the object to find the net force. The solution given here is more direct, but the solution that would have been obtain using the equations of motion would have been identical to this solution. The net force would have been found to be 300 Newtons. You can and, if time permits, probably should verify this. ]
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RESPONSE --> I felt pretty good about answering this problem self critique assessment: 3
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21:03:02 `q010. Solve the following without using any of the equations of motion. A net force of 5,000 Newtons acts on an automobile of mass 2,000 kg, initially at rest, through a displacement of 80 meters, with the force always acting parallel to the direction of motion. What velocity does the automobile obtain?
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RESPONSE --> Fnet = 5000 N, m = 2000 kg, v0 = 0, .ds = 80 m Find vf without motion formulas 'dW = Fnet * 'ds, 'dW = 5000 N * 80 m = 400000 Joules KE = 1/2 * m * v^2, 400000J = 1/2 * 2000kg * v^2 v^2 = 400000/1000 = 400 vf = 20 m/s confidence assessment: 3
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21:04:49 The We know that the net force does work `dWnet = Fnet * `ds = 5000 Newtons * 80 meters = 400,000 Joules. We know that the kinetic energy of the automobile therefore changes by 400,000 Joules. Since the automobile started from rest, its original kinetic energy KE0 was 0. We conclude that its final kinetic energy KEf must have been 400,000 Joules. Since KEf = 1/2 m vf^2, this is an equation we can solve for vf in terms of m and KEf, both of which we now know. We can first multiply both sides of the equation by 2 / m to obtain 2 * KEf / m = vf^2, then we can take the square root of both sides of the equation to obtain vf = +- `sqrt(2 * KEf / m) = +- `sqrt( 2 * 400,000 Joules / (2000 kg) ) = +- `sqrt( 400 Joules / kg). At this point we had better stop and think about how to deal with the unit Joules / kg. This isn't particularly difficult if we remember that a Joule is a Newton * meter, that a Newton is a kg m/s^2, and that a Newton * meter is therefore a kg m/s^2 * m = kg m^2 / s^2. So our expression +- `sqrt(400 Joules / kg) can be written +_`sqrt(400 (kg m^2 / s^2 ) / kg) and the kg conveniently divides out to leave us +_`sqrt(400 m^2 / s^2) = +- 20 m/s. We choose +20 m/s because the force and the displacement were both positive. Thus the work done on the object by the net force results in a final velocity of +20 m/s.
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RESPONSE --> I keep forgetting that the square root of anything can still be a positive number or a negative number
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21:20:29 `q011. If the same net force was exerted on the same mass through the same displacement as in the previous example, but with initial velocity 15 m/s, what would then be the final velocity of the object?
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RESPONSE --> Fnet = 5000 N, m = 2000 kg, v0 = 15 m/s, ds = 80 m Find vf without motion formulas 'dW = Fnet * 'ds, 'dW = 5000 N * 80 m = 400000 Joules KEv0 = 1/2 * 2000 * 15^2 = 225000 Joules 'dW = KEf - KE0, KEf = 'dW + KE0 KEf = 400000 + 225000 = 625000 Joules KEf = 625000 Joules KE = 1/2 * m * vf^2 vf^2 = 625000/(1/2 * 2000) = 625000/1000 = vf^2 = 625 vf = 25 m/s confidence assessment: 1
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21:23:20 Again the work done by the net force is still 400,000 Joules, since the net force and displacement have not changed. However, in this case the initial kinetic energy is KE0 = 1/2 m v0^2 = 1/2 (200 kg) (15 m/s)^2 = 225,000 Joules. Since the 400,000 Joule change in kinetic energy is still equal to the work done by the net force, the final kinetic energy must be KEf = KE0 + `dKE = 225,000 Joules + 400,000 Joules = 625,000 Joules. Since 1/2 m vf^2 = KEf, we again have vf = +- `sqrt(2 * KEf / m) = +-`sqrt(2 * 625,000 Joules / (2000 kg) ) = +-`sqrt(2 * 625,000 kg m^2/s^2 / (2000 kg) ) = +-`sqrt(625 m^2/s^2) = 25 m/s.
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RESPONSE --> It's a good thing I got a bunch of formulas written down from my previous work exposure even though I have 'dW = KEf + KE0 which is close to the KEf = KE0 + 'dKE in this case the 'dKE will equal 'dW self critique assessment: 2
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21:35:48 `q012. Solve without using the equations of motion: A force of 300 Newtons is applied in the direction of motion to a 20 kg block as it slides 30 meters across a floor, starting from rest, moving against a frictional force of 100 Newtons. How much work is done by the net force, how much work is done by friction and how much work is done by the applied force? What will be the final velocity of the block?
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RESPONSE --> F = 300 N, Ffrict = 100 N, m = 20 kg, 'ds = 30 m Find 'dWnetforce, 'dWfriction, 'dWappliedforce and vf W = F * 'ds = 300 * 30 = 9000 Joules W = Ffrict * 'ds = 100 * 30 = 3000 Joules W = Fnet * 'ds = (300 - 100) * 30 = 200 * 30 = 6000 Joules Wfnet = 6000 Joules Wfrict = 3000 Joules Wfapp = 9000 Joules KEvf = 1/2 * m * vf^2 = vf^2 = 6000/(1/2 * 20) = 6000/10 = 600 vf^2 = 600 vf = 24.49 m/s confidence assessment: 2
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21:38:39 The block experiences a force of 300 Newtons in its direction of motion and a force of 100 Newtons opposite its direction motion. It therefore experiences a net force of Fnet = 300 N - 100 N = 200 N. The work done by the net force is therefore `dWnet = 200 N * 30 m = 6000 Joules. The work done by the 300 Newton applied force is `dWapplied = 300 N * 30 m = 9000 Joules. The work done by friction is `dWfrict = -100 N * 30 m = -3000 Joules (note that the frictional forces in the direction opposite to that of the displacement). Note that the 6000 J of work done by the net force can be obtained by adding the 9000 J of work done by the applied force to the -3000 J of work done by friction. The final velocity of the object is obtained from its mass and final kinetic energy. Its initial KE is 0 (it starts from rest) so its final KE is KEf = 0 + `dKE = 0 + 6000 J = 6000 J. Its velocity is therefore vf = +- `sqrt(2 KEf / m) = `sqrt(12,000 J / (20 kg) ) = +-`sqrt( 600 (kg m^2 / s^2) / kg ) = +-`sqrt(600 m^2/s^2) = +- 24.5 m/s (approx.). We choose the positive final velocity because the displacement and the force are both in the positive direction.
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RESPONSE --> I guess I have to recognize the friction being opposite in magnitude even though when I computed the net I subtracted the friction because I knew it was against the applied force self critique assessment: 2
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