course Phy 121 I'm still working on the Force Vectors lab exercise. How do I find the lengths of the vectors? I know I use the 4cm per Newton scale. But how do I find the vectors in the first place? I know it's not the same as the length of the rubber bands. Does it have something to do with the angles that are clockwise from the positive x-axis?
......!!!!!!!!...................................
18:25:12 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
......!!!!!!!!...................................
RESPONSE --> `ds = 2m a = g = 9.8m/s^2 vo = 0m/s Since we only have three variables, we must use one of the equations of uniformly accelerated motion: `ds = vo*`dt + 1/2 a * `dt^2 2m = 0 + 1/2(9.8m/s^2)(`dt^2) `dt^2 = 0.41 s^2 `dt = 0.64 seconds
.................................................
......!!!!!!!!...................................
18:25:34 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
......!!!!!!!!...................................
RESPONSE --> I understand this problem.
.................................................
......!!!!!!!!...................................
18:27:21 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
......!!!!!!!!...................................
RESPONSE --> vAve = 12m/s `dt = 0.64s `ds = (vAve)(`dt) = (12m/s)(0.64s) = 7.7 meters
.................................................
......!!!!!!!!...................................
18:27:45 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
......!!!!!!!!...................................
RESPONSE --> I understand this problem.
.................................................
......!!!!!!!!...................................
18:34:48 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
......!!!!!!!!...................................
RESPONSE --> vo = 0m/s vAve = 6m/s `ds = 1.5m a = g = 9.8m/s^2 `ds = vo*`dt + 1/2(a)(`dt^2) `ds = 1/2(a)(`dt^2) 1.5m = 1/2(9.8m/s^2)(`dt^2) `dt^2 = 0.306s^2 `dt = 0.553 seconds `ds = vAve * `dt `ds = (6m/s)(0.553s) = 3.32 meters
.................................................
......!!!!!!!!...................................
18:35:01 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
......!!!!!!!!...................................
RESPONSE --> I understand this problem.
.................................................
......!!!!!!!!...................................
18:42:39 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
......!!!!!!!!...................................
RESPONSE --> Vertically: a = g = 9.8m/s `ds = 4m vo = 0m/s `ds = vo*`dt + 1/2(a)(`dt^2) `ds = 1/2(a)(`dt^2) 4m = 1/2(9.8m/s^2)(`dt^2) `dt^2 = 0.82s^2 `dt = 0.90 s vf =`dv = a * `dt = (9.8m/s^2)(0.9s) = 8.82m/s Horizontally: `ds horiz = 32m `dt = 0.9s vAve = `ds/`dt = 32m/0.9s = 35.6m/s = vf If the object moves 32 meters horizontally in 0.9 seconds, the average velocity is 35.6 m/s. The acceleration in the horizontal direction is uniform. Therefore, the velocity does not change. So the final velocity is equal to the average velocity.
.................................................
......!!!!!!!!...................................
18:43:23 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
......!!!!!!!!...................................
RESPONSE --> Ok. Had to think about this one. I tried to make it harder than it was.
.................................................
??s??h????Hi??W?assignment #016 016. `query 16 Physics I 06-27-2006
......!!!!!!!!...................................
20:48:16 Class notes #15 When a projectile rolls off a ramp with its velocity in the horizontal direction, why do we expect that its horizontal range `dx will be proportional to the square root of its vertical displacement `dy rolling down the ramp?
......!!!!!!!!...................................
RESPONSE --> `dx is proportional to sqrt(`dy) vo = 0m/s As an object falls kinetic energy is exerted. (KE = 1/2mv^2) From this information we can see that the vertical distance the object falls is proportional to the amount of KE exerted. (`dy is proportional to KE) As shown in previous problems we use the vertical displacement to solve for the distance and time during horizontal movement, and therefore solve for the horizontal velocity. So, when KE = 1/2mv^2, the horizontal velocity is equal to the square root of (2 KE/m). If the horizontal velocity is proportional to the square root of kinetic energy, it is also proportional to the vertical displacement, `dy.
.................................................
......!!!!!!!!...................................
20:48:34 ** Since the initial vertical velocity is zero the time of fall for a given setup will always be the same. Therefore the horizontal range is proportional to the horizontal velocity of the projectile. The horizontal velocity is attained as a result of vertical displacement `dy, with gravitational PE being converted to KE. PE loss is proportional to `dy, so the KE of the ball as it leaves the ramp will be proportional to `dy. Since KE = .5 m v^2, v is proportional to sqrt( KE ), therefore to sqrt(y). **
......!!!!!!!!...................................
RESPONSE --> My explanation isn't very clear. I should have mentioned that the horizontal range is proportional to the horizontal velocity because the initial velocit is zero. I probably also should have mentioned that as an object falls PE is converted to KE and so PE loss is proportional to KE exerted and therefore proportional to `dy.
.................................................
......!!!!!!!!...................................
20:51:01 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
......!!!!!!!!...................................
RESPONSE --> The initial velocity is zero, so the average velocity during vertical movement is equal to the final velocity. vf^2 = vo^2 + 2(a*`dy) vf = sqrt(2a`dy) Therefore, velocity is proportional to `dy and when we set up the equation for kinetic energy we can see more proportionalities: KE = 1/2mv^2 KE = 1/2m(sqrt 2a`dy)^2 KE = 1/2m(2a`dy) KE = m*a*`dy KE and `dy are proportional to each other.
.................................................
......!!!!!!!!...................................
20:51:29 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
......!!!!!!!!...................................
20:53:48 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
......!!!!!!!!...................................
RESPONSE --> Not all potential energy is converted to kinetic energy. Some is lost due to friction and other nonconservative forces. Therefore, the KE exerted is less than the PE that was stored.
.................................................
......!!!!!!!!...................................
20:54:40 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok. I did not mention the thermal energy, but I understand that it is another source of KE dissipation.
.................................................
......!!!!!!!!...................................
20:56:10 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
......!!!!!!!!...................................
RESPONSE --> First we must convert to standard units of velocity. (105km/hr)(1000m/km)(1hr/3600s) = 29.17m/s Now with mass and velocity, we can find the change in KE. We know that the final velocity is zero, so the final KE is also zero. Now we solve for the initial velocity. KE = 1/2mv^2 = 1/2(1250kg)(29.17m/s)^2 = 531805 Joules `dKE = 0 - 531805J = -531805 Joules The work done is equal and opposite to the KE. Therefore, `dW = -`dKE = 531805 Joules
.................................................
......!!!!!!!!...................................
20:56:35 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.
......!!!!!!!!...................................
RESPONSE --> I understand this problem.
.................................................
......!!!!!!!!...................................
20:57:25 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
......!!!!!!!!...................................
RESPONSE --> stiffness constant (k) = 440N/m PE = 25 J `ds = ? `dW = FAve * `ds The force exerted on the spring when it is relaxed is 0 Newtons. The force exerted on the spring when it is stretched is the product of the stretch constant and the distance it is stretched. F = k * `ds = N/m * m = N FAve = 0 + (k*`ds)/2 = (k*`ds)/2 `dW = (k*`ds)/2 * (`ds) = 1/2k*`ds^2 `dPE = `dw = 1/2k*`ds^2 25 Joules= 1/2(440N/m)(`ds^2) `ds^2 = 0.114 m^2 TAKE SQUARE ROOT `ds = +-0.338 meters
.................................................
......!!!!!!!!...................................
20:57:47 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.
......!!!!!!!!...................................
RESPONSE --> In the explanation I see that the force at position x is negative. Why is this?
.................................................
......!!!!!!!!...................................
20:58:01 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?
......!!!!!!!!...................................
RESPONSE --> I am not required to answer this question.
.................................................
......!!!!!!!!...................................
20:58:04 ** 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.. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
20:58:18 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?
......!!!!!!!!...................................
RESPONSE --> I am not required to answer this question.
.................................................
......!!!!!!!!...................................
20:58:21 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
20:58:32 Univ. 6.90 (6.78 10th edition) requires 10-25 watts / kg to fly; 70 g hummingbird 10 flaps/sec, work/wingbeat. Human mass 70 kg, 1.4 kW short period, sustain 500 watts. Fly by flapping?
......!!!!!!!!...................................
RESPONSE --> I am not required to answer this question.
.................................................
......!!!!!!!!...................................
20:58:36 ** A 70 gram = .070 kg hummingbird would require between .070 kg * 10 watts / kg = .7 watts = .7 Joules / second in order to fly. At 10 flaps / second that would be .07 Joules per wingbeat. A similar calculation for the 25 watt level shows that .175 Joules would be required per wingbeat. A 70 kg human being would similarly require 700 watts at 10 watts / kg, which would be feasible for short periods (possibly for several minutes) but not for a sustained flight. At the 25 watt/kg level flight would not be feasible. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"