course Phy 201 ????????{?????assignment #015015. `query 15
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18:48:00 Set 4 probs 1-7 If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?
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RESPONSE --> Momentum confidence assessment: 3
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18:48:05 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
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RESPONSE --> I understand. self critique assessment: 3
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18:48:45 What is the definition of the momentum of an object?
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RESPONSE --> mass*velocity confidence assessment: 3
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18:48:50 ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **
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RESPONSE --> I understand. self critique assessment: 3
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18:49:51 How do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?
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RESPONSE --> m=Fnet*'dt confidence assessment: 2
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18:50:09 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
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RESPONSE --> ok I put fNet, but it is average force. I understand. self critique assessment: 2
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18:50:33 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
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RESPONSE --> Not sure confidence assessment: 0
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18:53:08 ** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt **
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RESPONSE --> We see that the change in velocity equals a*'dt. We can then substitute a into the Force equation. Then, in place of a for the 'dv equation, we substitute (F/m), multiply both sides by m, and get the equation for momentum. self critique assessment: 2
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18:56:14 If you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies?
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RESPONSE --> m `dv = Fave `dt or aAve = (vf - v0) / `dt. IN this one, after finding aAve, we would have to multiply by mass to get the aAve. confidence assessment: 2
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18:56:21 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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RESPONSE --> I understand. self critique assessment: 3
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19:00:50 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
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RESPONSE --> We use the acceleration equation, then the final velocity equation (subbing in Fnet/m for a), then we use the change in KE formula to solve for work, which is 'ds*Fnet. I think this is the right progression anyway... confidence assessment: 1
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19:00:56 ** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **
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RESPONSE --> Okay, I understand. self critique assessment: 3
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19:01:38 What is kinetic energy and how does it arise naturally in the process described in the previous question?
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RESPONSE --> KE equals work. I don't know if that answers the quesiton, but... confidence assessment: 1
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19:01:57 ** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. **
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RESPONSE --> I see, it is not just ""work,"" but more specifically the ""work done by the net force."" self critique assessment: d2
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19:02:09 What forces act on an object as it is sliding up an incline?
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RESPONSE --> friction and gravity confidence assessment: 3
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19:02:17 ** Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force. Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion. The gravitational force is conservative; all other forces in the direction of motion are nonconservative. COMMON ERROR: The Normal Force is in the upward direction and balances the gravitational force. COMMENT: The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal. The normal force cannot balance the gravitational force if the incline isn't horizontal. Friction provides a component parallel to the incline and opposite to the direction of motion. **
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RESPONSE --> I understand. self critique assessment: 3
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19:55:25 For an object sliding a known distance along an incline how do we calculate the work done on the object by gravity? How do we calculate the work done by the object against gravity?
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RESPONSE --> Fg=m*g Wg=Fg*'ds I think the work done against gravity will be equal and opposite to the Wg above. confidence assessment: 2
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19:56:00 ** The gravitational force is m * g directly downward, where g is the acceleration of gravity. m * g is the weight of the object. If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of which is positive and/or negative. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. The precise value of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal. This force acts down the incline. If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase. This behavior is consistent with our experience of objects moving freely down inclines. If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative. In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines. The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **
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RESPONSE --> I understand. self critique assessment: 2
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20:02:13 For an object sliding a known distance along an incline how do we calculate the work done by the object against friction? How does the work done by the net force differ from that done by gravity?
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RESPONSE --> Work against friction=Ff*'ds The net force is the total gravitational force and the total frictional force that is acting on an object at a given time. confidence assessment: 3
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20:02:19 ** The work done against friction is the product of the distance moved and the frictional force. Since the force exerted by friction is always opposed to the direction of motion, the force exerted by the system against friction is always in the direction of motion so the work done against friction is positive. The net force on the system is sum of the gravitational component parallel to the incline and the frictional force. The work done by the net force is therefore equal to the work done by gravity plus the work done by the frictional force (in the case of an object moving up an incline, both gravity and friction do negative work so that the object must do positive work to overcome both forces; in the case of an object moving down an incline gravity does positive work on the system while friction, as always, does negative work on the system; in the latter case depending on whether the work done by gravity on the system is greater or less than the frictional work done against the system the net work done on the system may be positive or negative) **
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RESPONSE --> I understand. self critique assessment: 3
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20:02:45 Explain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length.
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RESPONSE --> It seems like I should be able to rationalize this, but I am having a hard time doing so. confidence assessment: 0
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20:03:01 ** In terms of similar triangles: The reason the approximation only works for small displacements is because the sides used on one triangle are not the same as the sides used on the other. From the triangle we see that the restoring force and the weight are at right angles, while the length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle and hence are not at right angles. For small angles the two long sides of the triangle are approximately equal so the discrepancy doesn't make much difference. For larger angles where the two long sides are significantly different in length, the approximation no longer works so well. In terms of components of the vectors: The tension force is in the direction of the string. The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement (just draw the picture). The vertical component of the tension force must be equal to the weight of the pendulum, since the pendulum is in equilibrium. If the displacement is small compared to the length the vertical component of the tension force will be very nearly equal to the tension force. So the previous statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement' can be replaced by the statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the weight of the pendulum as the length of the pendulum to the horizontal displacement. **
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RESPONSE --> This is complicated, but I sort of understand the general idea. self critique assessment: 1
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???????€????}?assignment #018 018. `query 18 Physics I 06-11-2007
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21:37:04 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
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RESPONSE --> confidence assessment:
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???????????assignment #018 018. `query 18 Physics I 06-11-2007
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21:54:33 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
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RESPONSE --> Well, you can't find horizontal and vertical projectile ranges at the same time; they have to found separately. We generally can use the vf^2=vo^2+2a'ds formula to find final velocityof the vertical fall. Then, we can use the displacement and the velocities to find 'dt. When we have 'dt, we can use this same time for the horizontal velocity. We could use the formula vf^2=vo^2+2a'ds rearranged to solve for 'ds in order to find the horizontal range. confidence assessment: 2
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21:57:42 ** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
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RESPONSE --> I undersrand. self critique assessment: 3
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22:01:59 Query class notes #17 Why do we expect that in a collision of two objects the momentum change of each object must be equal and opposite to that of the other?
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RESPONSE --> Since momentum has to do with mass and velocity I am assuming the masses and velocities will remain what they were from beginning to end, and since one object is coming from direction and the other object frmo another direction, they will be opposite momentums impacting. confidence assessment: 2
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22:02:34 **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding object exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
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RESPONSE --> I get the general idea. self critique assessment: 2
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22:14:34 What are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
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RESPONSE --> Momentum has to do with mass and velocity, and since the objects are moving and then colliding, there will be an initial and a final velocity for each object. confidence assessment: 3
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22:15:17 ** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
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RESPONSE --> I understand that the totals for the finals and for the initials have to be equal and opposite. self critique assessment: 2
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22:45:20 `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
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RESPONSE --> I think most if not all of the KE will turn in to thermal energy. 95 km/hr=26.4 m/s KEo=.5mv^2 =.5(7650kg)(26.4m/s)^2 =265,000J self critique assessment: 2
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22:46:03 There is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J. This KE is practially all converted to thermal energy.
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RESPONSE --> I see that I should have multiplied KEo by 2 to get the total KE. Otherwise, I understand. self critique assessment: 2
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22:59:45 `1Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
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RESPONSE --> vo=1.7m/s 'ds=45m I am stuck... self critique assessment: 0
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23:05:42 **GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
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RESPONSE --> 'dKE+'dPE+'dWnoncons=0 I understand how the appropriate equations for each of the above parts of the equation are filled in and then mass cancels, making it unimportant to know mass. I also understand how the final answer was derived from simply filling in the known variables. The only thing I am still unsure about is what (y2-y1) is exactly... I understand it is filled in for PE, but I don't know what those variables represent in this equation. self critique assessment: 2
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23:05:45 Univ. 7.74 (7.62 in 10th edition). 2 kg pckg, 53.1 deg incline, coeff kin frict .20, 4 m from spring with const 120 N/m. Speed just before hitting spring? How far compressed? How close to init pos on rebound?
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RESPONSE --> confidence assessment:
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23:05:48 ** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package. The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline. The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx.. The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx.. Friction acts in the direction opposite motion, up the incline in this case. If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is 13.4 N * 4 m = 54 Joules approx. Just before hitting the spring we therefore have .5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s. If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx.. However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed. As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have `dKE + `dPE + `dWnoncons = 0 so -54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation 60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8 meaning 1.07 m or -.8 m (see previous note on units). We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE. If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain `dPE + `dKE + `dWnoncons = 0 so -.5 k x^2 + Fparallel * xMax + 0 + 33 N * xMax = 0 or -.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0 We obtain 18 N * xMax = 72 N m, approx., so that xMax = 72 N m / (18 N) = 4 meters, approx.. This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
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RESPONSE --> self critique assessment:
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23:05:51
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RESPONSE --> self critique assessment:
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