#$&* course Phy 121 Please disregard the open query i just submitted it was this open QA, im submitting the open QA now with the proper title 013. Energy*********************************************
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Given Solution: We know the initial velocity v0 = 0 and the displacement `ds = 20 meters. We have the information we need to determine the acceleration of the object. Once we find that acceleration we can easily determine its final velocity vf. We first find the acceleration. The object is subjected to a net force of 40 Newtons and has mass 10 kg, so that will have acceleration a = Fnet / m = 40 Newtons / 10 kg = 4 m/s^2. We can use the equation vf^2 = v0^2 + 2 a `ds to see that vf = +- `sqrt( v0^2 + 2 a `ds ) = +- `sqrt ( 0 + 2 * 4 m/s^2 * 20 meters) = +-`sqrt(160 m^2 / s^2) = +-12.7 m/s. The acceleration and displacement have been taken to be positive, so the final velocity will also be positive and we see that vf = + 12.7 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q002. Find the value of the quantity 1/2 m v^2 at the beginning of the 20 meter displacement, the value of the same quantity at the end of this displacement, and the change in the quantity 1/2 m v^2 for this displacement. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ½ (10kg)(0)= 0 1/2 (10kg)(12.7m/s) = 63.5kg*m/s^2 (N) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to square the velocity ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q003. Find the value of the quantity Fnet * `ds for the present example, and express this quantity in units of kg, meters and seconds. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The net force is 40 newtons, the displacement 20 meters. Fnet (40N) * `ds (20m) = 800 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Fnet = 40 Newtons and `ds = 20 meters, so Fnet * `ds = 40 Newtons * 20 meters = 800 Newton meters. Recall that a Newton (being obtained by multiplying mass in kg by acceleration in m/s^2) is a kg * m/s^2, so that the 800 Newton meters can be expressed as 800 kg m/s^2 * meters = 800 kg m^2 / s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The quantities are the same, therefore Fnet*ds = 1/2mv^2 (final) - 1/2mv^2 (initial) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = 10kg Net force = 40 N `ds = 20m A = 4m/s^2 Vo = 9m/s Vf = `sqrt [(9m/s)^2 + 2 (4m/s^2)(20m) Vf = `sqrt 81m^2/s^2 + 160m^2/s^2 = `sqrt 241m^2/s^2 Vf = 15.5m/s `KE = ½(10kg)(15.5m/s)^2 - ½ (10kg)(9m/s)^2 = = 1201.25N - 405N = 796.25 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.] STUDENT COMMENT: I rounded to 15.5m/s instead of using sqrt(241). Will use the more accurate value in the future. INSTRUCTOR RESPONSE As long as you round to the appropriate number of significant figures, you're OK. However in situations where you're going to be squaring the result anyway, you introduce less roundoff error if you leave it in the radical form. In these cases radical form is simply more convenient. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = 20 N M = 4kg `ds = 10m Vo = 0 A = 5m/s^2 Vf = 10m/s Fnet (20N) * `ds (10m) = 200J ½ (4kg)(10m/s)^2 - 0 = 200J These random numbers I used tell us that the fnet*`ds always equals 1/2mv^2 - 1/2mv^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t able to properly calculate it in this way with just the units but I see what you did here. ------------------------------------------------ Self-critique rating: