course PHY 121 I will take the test tomorroww(Friday) but I am not counting on passing when I cannot do the problems with my notes. I am just not understanding the KE and PE and formulas to calculate the problems and all their steps. It is not sinking in. șy̦ƠSǡassignment #002
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11:03:59 1) When masses of 35, 70 and 105 grams are hung from a certain rubber band its respective lengths are observed to be 39, 46, and 53 cm. What are the x and y components of the tension of a rubber band of length 50.6 cm if the x component of its length if 14.02232? What horizontal force, when added to this force will result in a total force of magnitude 150 grams?
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RESPONSE --> m = 35g L=39cm m=70g L=46cm m=105g L=53cm 50.6cm/14.02232 = 3.609 ?
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11:14:21 STUDENT SOLUTION WITH INSTRUCTOR COMMENTS INDICATED BY ** I am assuming that you plug in what you know into the equation of the line. Y = .2x + 32 ** If y represents the mass and x the length that approach would give you the right mass--good thinking, though the equation would be more like y = 5 x - 150. Your equation looks reasonably good if y is length and x is supported mass. However you do need to use the force, which will be the force exerted by gravity on the mass. Force and mass are not the same thing. In general you will use the graph to determine the force corresponding to a given length. If there is significant curvature to the graph the linear model won?t work very well. In that case you could just draw a smooth curve to fit the data and read your forces from the curve. ** When the length is 50.6 cm the corresponding weight is 93 grams and when the length is 14.02232 cm the weight is ?89.9 g, but I don?t think weight can be negative???? ** Plug the rubber band length into the force relationship or read the force from your graph. If your graph is of supported mass vs. length you need to find the force. In this case you have a length of 50.6 cm, which corresponds to a supported mass of about 93 grams. This corresponds to a force of about .093 kg * 9.8 m/s^2 = .91 Newtons. The direction of the rubber band is arctan(y/x). Using the Pythagorean Theorem with length 50.6 cm and x component 14 cm we get y component about 48.6 cm, so the direction of the rubber band is arctan(48.6 / 14) = 74 deg. The angle at which the tension acts is parallel to the rubber band, so the tension also acts at 74 deg. The x and y components of the force are therefore Fx = .91 N * cos(74 deg) = .25 N and Fy = .91 N * sin(74 deg) = .87 N. If a horizontal force Fhoriz is added then the x component becomes Fx = .25 N + Fhoriz and the magnitude of the resulting force is | F | = sqrt(Fx^2 + Fy^2) = sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2). If this force is 150 grams (or, converting this to a force, 1.47 N) then we have sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2) = 1.47 N. Squaring both sides we get .0625 N^2 + .5 N * Fhoriz + Fhoriz^2 + .76 N^2 = 2.16 N^2. This is quadratic in Fhorz and rearranges to Fhoriz^2 + .5 N * Fhoriz ? 2.1 N^2 = 0. Using the quadratic formula we get Fhoriz = (-.5 N +- sqrt( (.5 N)^2 ? 4 * 1 * (-2.1 N^2) ) / (2 * 1) = (-.5 N +- 2.8 N) / 2, giving us two forces: Fhoriz = (-.5 N + 2.8 N ) / 2 = 1.15 N and Fhoriz = (-.5 N ? 2.8 N) / 2 = -1.65 N.**
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RESPONSE --> I do not see where you are getting 93 grams for supported mass? Is the .5N * Fhoriz the same as .5a(acceleration)? Even with notes if front of me I do not understand how to follow this flow of reasoning and formulas.
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11:19:04 Intuitively the PE increases as the ball rises, so that KE decreases. Since we know that the KE of the rising ball is 0 at the top of the arc we can find the initial KE by finding the PE increase. As the ball falls, PE decreases and KE increases. Between initial and final position the ball has a small net downward displacement, hence a small net loss of PE, so it has a small net gain in KE. PE changes are easily found by multiplying the weight of the ball by the displacement. The detailed solution that follows is based on these concepts, but uses the work-energy theorem in a rigorous fashion: The work done by gravity on the ball is equal to the product of the force exerted by gravity, which is .3 kg * 9.8 m/s^2 = 2.94 N downward, and the displacement, which is 4.68 m ? 4.1 m = .58 m upward. This work is therefore -2.94 N * .58 m = -1.8 Joules, approx.. The work done by the ball against gravity is +1.8 J, equal and opposite to the work done by gravity on the ball. The work done against gravity is the change in the PE of the ball. Assuming no dissipative forces we have `dPE + `dKE = 0, so the KE change during this upward displacement is `dKE = -`dPE = -1.8 J. Since the final KE is 0, we have `dKE = KEf - KE0 = -1.8 J so that KE0 = 1.8 J + KEf = 1.8 J + 0 = 1.8 J. From initial to final position altitude changes by 4.06 m ? 4.1 m = -.04 m. Thus PE changes by -.04 m * 2.94 N = -.12 J. Since KE0 = 1.8 J and `dKE + `dPE = 0 we have `dKE = -`dPE = +.12 J so that KEf ? KE0 = .12 J and KEf = KE0 + 1.2 J = 1.8 J + .12 J = 1.92 J. The final velocity of the ball therefore satisfies .5 m vf^2 = KEf so that vf = +-sqrt(2 KE / m) = +- sqrt(1.92 J / (.3 kg) ) = +- sqrt( 6.4 m^2/s^2) = +- 2.5 m/s, approx.. Since final velocity is downward, our previous implicit choice of upward as the positive direction tells us that the final velocity is -2.5 m/s. The work done by the ball against gravity is equal and opposite to its change in KE.
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RESPONSE --> dPE = weight * ds The work against is -1.8J which is opposite of +1.8J of work done by the ball. .5m vf^2 = KEF= +-sqrt(2KE/m)
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11:22:45 3) Give an example of a situation in which you are given v0, a, and ?ds, and reason out all possible conclusions that could be drawn from these three quantities assuming uniform acceleration.
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RESPONSE --> I can calculate final velocity by vf^2 = vO^2 + 2 a ds from this I can get Fnet ds = 1/2mvf^2 - 1/2mv0^2 I can find the change in time by ds = vo dt + 1/2 a dt^2
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11:22:58 ** Your solution should show the use of the fourth equation of motion vf^2 = v0^2 + 2 a `ds to find vf. You then have the initial and final velocities, which since acceleration is uniform can be averaged to give you the average velocity vAve and the change `dv in velocity.. Dividing displacement `ds by average velocity we obtain `dt.
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11:43:52 4) A ball reaches a ramp of length 74 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 5.2 seconds. Its velocity at the end of the ramp is 7.46154 cm/s. What is its acceleration on the ramp?
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RESPONSE --> vf=7.46154cm/s vo = ? L = 74cm dt = 5.2s 74cm = (vo + 7.46154cm/s)/2 * 5.2s (74cm/5.2s) * 2 = vo + 7.46154cm/s 28.46cm/s - 7.46154cm/s = 20.998cm/s This is my initial velocity 7.46154cm/s^2 = 20.998cm/s 7.46154cm/s = 20.998cm/s + a 5.2s (7.46154cm/s - 20.998cm/s )/5.2s = -2.603cm/s^2 a = -2.603cm/s^2
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11:44:07 STUDENT SOLUTION: I found the average velocity to be 74 cm / (5.2 s) = 14.23 cm/s. Since vAve = (vf + v0) / 2 we can easily solve for v0, obtaining v0 = 2 vAve ? vf = 2 * 14.23 cm/s ? 7.46 cm/s = 21 cm/s. Having vf and v0 we easily find that `dv = vf ? v0 = 7.46 cm/s ? 21 cm/s = -13.54 cm/s, and therefore a = `dv / `dt = -13.54 cm/s / (5.2 s) = ?2.6 cm/s^2
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11:55:47 5) A ball starting from rest rolls 10 cm down an incline on which its acceleration is constant, requiring .83 seconds to cover the distance. It then rolls onto a second incline 41 cm long on which its acceleration is 14 cm/s^2. How much time does it spend on the second incline and what is its acceleration on the first?
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RESPONSE --> v0 = 0 ds = 10cm dt = .83s 10cm/.83s = 12.05cm/s(vAve) 12.05cm/s = a .83 12.05cm/s / .83s = 14.52cm/s^2 acceleration on first ramp vo on second ramp is vf on first 12.05cm/s is vf on first ramp 41cm = 12.05cm/s * dt + .5 (14cm/s^2) * dt^2 41cm - 12.05cm/s*dt = 7dt^2 28.95cm/s / 7cm/s^2 = dt dt = 4.14s on second ramp
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11:57:57 STUDENT SOLUTION: Ramp 1 I found the average velocity vAve = 10 cm / (.83 s) = 12 cm/s, approx.. Since v0 = 0 we have vf = 2 * vAve = 24 cm/s, approx. Acceleration on this incline is a = `dv / `dt = 24 cm/s / (.83 s) = 29 cm/s^2 approx. Ramp 2 I plugged in what I knew to find the unknowns, using v0 = 24 cm/s (the final velocity on the first incline is the init vel on the second), a = 14 cm/s^2 and `ds = 41 cm. I obtained vf = sqrt( v0^2 + 2 a `ds) = sqrt( (24 cm/s)^2 + 2 * 14 cm/s^2 * 41 cm) = 41 cm/s, approx.. Average velocity on the second incline is therefore (41 cm/s + 24 cm/s) / 2 = 32.5 cm/s, and the time required is `dt = `ds / vAve = 41 cm / (32.5 cm/s) = 1.3 sec, approx.
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RESPONSE --> I did not multiply the average velocity by 2. I found the 12cm/s for final velocity and used that figure to calculate acceleration on the first ramp which through my calculations off.
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12:03:44 6) A cart of mass 1.7 kg coasts 30 cm down an incline at 4 degrees with horizontal. Assuming Ffr is .042 times the normal force and other nongravitational forces parallel to the incline are negligible. * What is the component of the carts weight parallel to the incline? * How much work does this force do as the cart rolls down the incline? * How much work does the net force do as the cart rolls down the incline? * Using the definition of KE, determine the velocity of the cart after coasting the 30 cm assuming it initial velocity is zero.
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RESPONSE --> 1.7kg * 9.8 = 16.6N 16.6N -.41N = 16.19N 16.19N * 30cm = 485.7N * cm 1.7kg cos(4) = 1.7 1.7kg sin(4) = .12
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12:04:48 The weight of the cart is 1.7 kg * 9.8 m/s^2 = 16.7 N, approx.. The incline makes an angle of 4 degrees with horizontal, so if the positive x axis is directed up the incline the weight will be at angle 270 deg ? 4 deg = 266 deg with the positive x axis and the x and y components of the weight will be x comp of weight = 16.7 N * cos(266 deg) = -1.2 N, approx., and y comp of weight = 16.7 N * sin(266 deg) = -16.6 N approx. The x component is parallel and the y component perpendicular to the incline. The normal force exerted by the incline is the elastic reaction to the y component of the weight and is +16.6 N. The parallel component of the gravitational force is in the direction of the displacement down the incline so the work done by this component on the cart is positive. We get work by parallel component of weight = 1.2 N * .30 m = .36 Joules, approx.. The frictional force is .042 times the normal force, or frictional force = .042 * 16.6 N = .7 N, approx., directed opposite to the displacement (up the incline). So the net force is net force = parallel component of gravitational force + frictional force = -1.2 N + .7 N = -.5 N, or .5 N down the incline. The work done by this force is .5 N * .3 m = .15 J. The KE of the cart after coasting down the incline from rest is by the work-energy theorem equal to the work done by the net force. So we have .5 m vf^2 = KE, so that vf = +- sqrt(2 KE / m) = +- sqrt( 2 * .15 J / (1.7 kg) ) = +- sqrt( .16 m^2/s^2) = +- .4 m/s, approx. Using the sign conventions specified by our choice of the upward direction as the positive x direction, the velocity will be -.4 m/s.
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12:14:45 7) A ball mass 8 kg rolls off the edge of a ramp with a horizontal speed of 70 cm/s. * What is its KE as it rolls off the ramp? * How much work does gravity do on the ball as it falls 22 cm? * What will be its KE after falling 22 cm? *How much of this KE is accounted for by its horizontal velocity and how much by its vertical velocity? * What then is its vertical velocity at this point?
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RESPONSE --> KE = 1/2 m v^2 KE = 1/2(8kg)(70cm/s)^2 KE = 19600kg cm dw = f * ds dw = 9.8 * 22cm dw = 215.6N 19600 - 215.6 = 19384.40N
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12:16:45 We have KE = .5 m v^2 = .5 * 8 kg * (.70 m/s)^2 = 2 Joules, approx.. As the ball falls 22 cm gravity does work `dWgrav = 8 kg * 9.8 m/s^2 * .22 m = 17 Joules, approx.. So after the 22 cm fall we will have KE = 2 J + 17 J = 19 J. There is no net force in the horizontal direction so the horizontal velocity will be unchanged and its horizontal KE will be 2 J, leaving 17 J of vertical KE. The vertical velocity will therefore be vertical velocity = +-sqrt(2 * vertical KE / m) = +- sqrt( 2 * 17 J / (8 kg)) = +-2.1 m/s.
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RESPONSE --> I did not see the .70, I read 70m/s I calculated the 8kg * 9.8m/s^2 but I don't think I multiplied it by 22cm
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