OpenQuery17

course PHY 121

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017. `query 17*********************************************

Question: `qprin phy and gen phy 6.33: jane at 5.3 m/s; how high can she swing up that vine?

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Your solution: `dKE + `dPE = 0. Jane’s KE during movement = .5(m)(v^2), and she will eventually come to a stop (when she starts to swing back down) when KE = 0. So `dKE = 0- (.5)(m)(v^2). PE = m *g* `dy.

So [0- (.5)(m)(v^2)] + [m *g* `dy] = 0. We want to get `dy (how high she can swing) alone.

[0 - .5mv^2]/m*g = `dy. The m’s cancel each other out, leaving 0 - .5v^2/ g = -`dy.

[.5(5.3m/s^2)] / (9.8m/s^2) = -`dy.

`dy = 1.4m

confidence rating:

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Given Solution: Jane is going to convert her KE to gravitational PE. We assume that nonconservative forces are negligible, so that `dKE + `dPE = 0 and `dPE = -`dKE.

Jane's KE is .5 M v^2, where M is her mass. Assuming she swings on the vine until she comes to rest at her maximum height, the change in her KE is therefore

`dKE = KEf - KE0 = 0 - .5 M v0^2 = - .5 M v0^2, where v0 is her initial velocity.

Her change in gravitational PE is M g `dy, where `dy is the change in her vertical position. So we have

`dKE = - `dPE, or

- 1/2 M v0^2 = - ( M g `dy), which we solve for `dy (multiply both sides by -1, divide both sides by M g) to obtain

`dy = v0^2 / (2 g) = (5.3 m/s)^2 / (2 * 9.8 m/s^2) = 1.43 m.

STUDENT QUESTION: I’m confused as to where the 2 g came from

INSTRUCTOR RESPONSE: You are referring to the 2 g in the last line.

We have in the second-to-last line

- 1/2 M v0^2 = - ( M g `dy). Dividing both sides by - M g, and reversing the right- and left-hand sides, we obtain

`dy = - 1/2 M v0^2 / (M g) = 1/2 v0^2 / g = v0^2 / (2 g).

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Self-critique (if necessary): OK – but that was rough.

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Self-critique rating:

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Question: `qprin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball

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Your solution: `dKE = -`dPE. And PE = .5kx^2. So PE = .5(950N/s)(.150m)^2 = 10.7Joules. Based on `dKE = -`dPE, the `dKE must be 10.7J. To find the speed, KE = .5(m)(v^2).

10.7J = .5(.3kg)v^2.

10.7J = 0.15 (v^2)

71.3 = v^2, taking square root, v = 8.4m/s. To find the maximum altitude, we have to somehow use the only equation I can think of that uses `dy, which is PE = g*m*`dy. We learned in the previous problem that [0- (.5)(m)(v^2)] + [m *g* `dy] = 0, and we want to isolate `dy. This leaves:

-.5v^2/g = `dy.

.5(8.4m/s)^2 = -35.3m.

confidence rating:

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Given Solution: We being with a few preliminary observations:

• We will assume here that the gravitational PE of the system is zero at the point where the spring is compressed. In this situation we must consider changes in both elastic and gravitational PE, and in KE.

• We also observe that no frictional or other nonconservative forces are mentioned, so we assume that nonconservative forces do no work on the system.

• It follows that `dPE + `dKE = 0, so the change in KE is equal and opposite to the change in PE.

The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 10.7 J.

Since the ball is moving in the vertical direction, between the release of the spring and the return of the spring to its equilibrium position, the ball has a change in gravitational PE as well as elastic PE.

• The change in elastic PE is -10.7 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +.44 J.

• The total change in PE is therefore -10.7 J + 4.4 J = -10.3 J.

Summarizing what we know so far:

• Between release and the equilibrium position of the spring, `dPE = -10.3 J

During this interval, the KE change of the ball must therefore be `dKE = - `dPE = - (-10.3 J) = +10.3 J.

Intuitively, the ball gains in the form of KE the 10.3 J of PE lost by the system.

The initial KE of the ball is 0, so its final KE during its interval of contact with the spring is 10.3 J. We therefore have

• .5 m v^2 = KEf so that

• vf=sqrt(2 KEf / m) = sqrt(2 * 10.3 J / .30 kg) = 8.4 m/s.

To find the max altitude to which the ball rises, we consider the interval between release of the spring and maximum height.

• At the beginning of this interval the ball is at rest so it has zero KE, and the spring has 10.7 J of elastic PE.

• At the end of this interval, when the ball reaches its maximum height, the ball is again at rest so it again has zero KE. The spring also has zero PE, so all the PE change is due to the gravitational force encountered while the ball rises.

• Thus on this interval we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. Since the spring loses its 10.7 J of elastic PE, the gravitational PE must increase by 10.7 J.

• The change in gravitational PE is equal and opposite to the work done on the ball by gravity as the ball rises. The force of gravity on the ball is m g, and this force acts in the direction opposite the ball's motion. Gravity therefore does negative work on the ball, and its gravitational PE increases. If `dy is the ball's upward vertical displacement, then the PE change in m g `dy.

• Setting m g `dy = `dPE we get

`dy = `dPE / (m g)

= 10.7 J / ( .30 kg * 9.8 m/s^2)

= 10.7 J / (2.9 N) = 10.7 N * m / (2.9 N) = 3.7 meters.

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Self-critique (if necessary): I tried to make it too complicated at the end; I could have just used `dPE = mg`dy instead of trying to work in a relationship to KE. From `dPE = mg*`dy, I could have divided to get `dPE/mg = `dy, leaving (10.7J)/(.3*9.8) = 3.7m.

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Self-critique rating:"

&#Your work looks good. Let me know if you have any questions. &#