#$&* course Phy 121 11/11 10 013. Energy
.............................................
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): When I took the sqrt of the number, my solution rounded to 12.6, not 12.7, does this really matter? ------------------------------------------------ Self-critique rating:3
.............................................
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): The rounding of my vf is different from the given solution, which gave me a slightly different answer here. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: Since the previous problem gave us Fnet = 40 N and `ds = 20 m, the quantity 40 kg * m/s^s * 20 m = 800 kg * m^2/s^2 which is 800 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They give you the same solution. Fnet * `ds solves for `dKE, while (1/2 m v^2) at a given velocity solves for KE. In this case, the way the questions were set up, we found the `dKE for both, which gave us the same answer (I rounded a number differently though and got a slightly different answer). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The change in the quantity Fnet * `ds is 800 kg m^2 / s^2 and the change in 1/2 m v^2 is 800 kg m^2 / s^2. The quantities are therefore the same. This quantity could also be expressed as 800 Newton meters, as it was in the initial calculation of the less question. We define 1 Joule to be 1 Newton * meter, so that the quantity 800 Newton meters is equal to 800 kg m^2 / s^2 and also equal to 800 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. Suppose that all the quantities given in the previous problem are the same except that the initial velocity is 9 meters / second. Again calculate the final velocity, the change in (1/2 m v^2) and Fnet * `ds. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = 40 Newtons, `ds = 20 m, and m = 10 kg a = 40 Newtons / 10 kg = 40 kg * m/s^2 / 10 kg = 4 m/s^2 vf^2 = v0^2 + 2 a `ds = (9 m/s)^2 + 2 * 4 m/s^2 * 20 m = 81 m^2/s^2 + 160 m^2/s^2 = 241 m^2/s^2 vf = sqrt(241 m^2/s^2) = 15.5 m/s KE = .5 m v^2 = .5 * 10 kg * (9 m/s)^2 = .5 * 10 kg * 81 m^2/s^2 = 405 Joules KE = .5 m v^2 = .5 * 10 kg * (15.5 m/s)^2 = .5 * 10 kg * 240.3 m^2/s^2 = 1201.5 Joules Therefore `dKE = KEf - KE0 = 1201.5 J - 405 J = 796.5 Joules `dKE = Fnet * `ds = 40 N * 20 m = 40 kg * m/s^2 * 20 m = 800 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I guess I need to start leaving it in the sqrt form from now on as well, since I was rounding it off. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: I am going to let Fnet = 80 Newtons, mass = 20 kg, and displacement = 40 m. I am going to say that v0 = 0 m/s. a = Fnet / m = 80 Newtons / 20 kg = 4 m/s^2 vf^2 = v0^2 + 2 a `ds = 0 + 2 * 4 m/s^2 * 40 m = 320 m^2/s^2, therefore vf = sqrt(320 m^2/s^2) = 17.9 m/s initial 1/2 m v0^2 = 1/2 * 20 kg * (0 m^2/s^2)^2 = 0 kg m^2/s^2 final 1/2 m vf^2 = 1/2 * 20 kg * (17.9 m/s)^2 = 3200 kg m^2/s^2 Therefore the change in the quantities of initial and final 1/2 m v^2 = 3200 kg m^2/s^2 Now we need to see if this equals the solution we get to Fnet * `ds = 80 Newtons * 40 m = 3200 kg m^2/s^2 Yes, these two results are equal. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dWnet = Fnet * `ds = 12 Newtons * 100 m = 1200 kg m^2/s^2 In order to find the value for the v in this next equation, we must find the a. We know since it started from rest, v0 = 0 a = Fnet / m = 12 Newtons / 48 kg = .25 m/s^2 Therefore vf^2 = v0^2 + 2 a `ds = 0 + 2 * .25 m/s^2 * 100 m = 50 m^2/s^2, therefore vf = sqrt(50 m^2/s^2) = 7.1 m/s KE0 = 1/2 m v0^2 = 1/2 * 48 kg * (0 m/s)^2 = 0 m^2/s^2 KEf = 1/2 m vf^2 = 1/2 * 48 kg * (7.1 m/s)^2 = 1209.8 kg m^2/s^2 Therefore the change in KE is approximately 1200 kg m^2/s^2 The answers are not the exact same, but when rounded, I do believe it is close enough to count. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If `dKE is equal to `dWnet, then we should be able to find the answers from this. KE0 = 1/2 * 200 kg * (5 m/s)^2 = 100 kg * 25 m^2/s^2 = 2500 Joules KEf = 1/2 * 200 kg * (10 m/s)^2 = 100 kg * 100 m^2/s^2 = 10000 Joules `dKE = KEf - KE0 = 10000 kg m^2/s^2 - 2500 kg m^2/s^2 = 7500 Joules THerefore the `dWnet = 7500 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The work done by the net force is equal to the change in the KE of the object. The initial kinetic energy of the object is KE0 = 1/2 m v0^2 = 1/2 (200 kg) (5 m/s)^2 = 2500 kg m^2/s^2 = 2500 Joules. The final kinetic energy is KEf = 1/2 m vf^2 = 1/2 (200 kg)(10 m/s)^2 = 10,000 Joules. The change in the kinetic energy is therefore 10,000 Joules - 2500 Joules = 7500 Joules. The same answer would have been calculated calculating the acceleration of the object, which because of the constant mass and constant net force is uniform, the by using the equations of motion to determine the displacement of the object, the multiplying by the net force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q009. Answer the following without using the equations of uniformly accelerated motion: If the 200 kg object in the preceding problem is uniformly accelerated from 5 m/s to 10 m/s while traveling 50 meters, then what net force was acting on the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can use the `dWnet that we found in the last problem as 7500 Joules to find the Fnet since `dWnet = Fnet * `ds Fnet = `dWnet / `ds = 7500 Joules / 50 m = 150 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. ] STUDENT COMMENT: i dont understand why we couldnt use accel. and mass. INSTRUCTOR RESPONSE You need to learn how to solve problems based on energy, simply because acceleration isn't always constant and you don't always have the information required to find acceleration. This is why you are asked to solve the problem using energy considerations. STUDENT COMMENT: i understand how to sole the problem but how do you find the acceleration given the velocities and mass? INSTRUCTOR RESPONSE: The acceleration is not necessary to solve this problem. It can be solved using energy considerations only. If you were to assume a time interval, then you could find the acceleration and the net force. Different time intervals would give you different accelerations and different net forces, as well as different displacements; however the work done would be the same in every case. The work done by the net force depends only on the mass and the initial and final velocities. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dWnet = Fnet * `ds = 5000 Newtons * 80 m = 400000 Joules If we know the v0 is 0, then we can easily use the 400000 Joules for our KEf to find the velocity that is obtained. KEf = 1/2 m vf^2 vf^2 = KEf / 1/2 m = 400,000 J / 1/2 * 2,000 kg = 400 J/kg, therefore vf = sqrt(400 J/kg) = +-20 m/s. The velocity would be positive, so vf = 20 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well from the last problem, we know that the `dKE = 400,000 Joules We would know that the KE0 = 1/2 * 2,000 kg * (15 m/s)^2 = 225,000 Joules Therefore with this given information, we would take KEf = `dKE + KE0 = 400,000 Joules - 225,000 Joules = 625,000 Joules vf^2 = `dKE / 1/2 m = 625,000 Joules / 1/2 * 2,000 kg = 625 Joules/kg, therefore vf = sqrt(625 Joules/kg) = +-25 m/s. In this case, the vf is +, therefore making vf = 25 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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 (2000 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am guessing the net force has no concern with positive or negative, just the fact that a force was done. Therefore Fnet = 300 Newtons + 100 Newtons = 400 Newtons In this case, `dWnet = Fnet * `ds = 400 N * 30 m = 12,000 Joules `dWfrict = 100 Newtons * 30 m = 3,000 Joules `dWapplied = 300 Newtons * 30 m = 9,000 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did this wrong. In earlier examples, sometimes the friction was observed as negative, and sometimes not, and I thought that those equations had the same senarios and could not figure out the difference. ------------------------------------------------ Self-critique rating:3