qa172

#$&*

course Phy 231

4-6-11 about 1:30 pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

017. `query 17

ANSWERS/COMMENTARY FOR QUERY 17

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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:

1/2m*v0^2= mgdy

½ v0^2= gdy

,v0^2/2g=dy

(5.3m/s)^2/2(9.8m/s^2)= 1.43m

confidence rating #$&*:

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Given Solution:

Outline of solution:

Jane has KE. She goes higher by increasing her gravitational PE.

Her KE is 1/2 m v_0^2, where m is her mass and v0 is her velocity (in this case, 6.3 m/s^2). If she can manage to convert all her KE to gravitational PE, her KE will decrease to 0 (a decrease of 1/2 m v0^2) and her gravitational PE will therefore increase by amount 1/2 m v_0^2.

The increase in her gravitational PE is m g `dy, where m is again her mass and `dy is the increase in her altitude.

Thus we have

PE increase = KE loss

In symbols this is written

m g `dy = 1/2 m v0^2.

The symbol m stands for Jane's mass, and we can also divide both sides by m to get

g `dy = 1/2 v0^2.

Since we know g = 9.8 m/s^2 and v0 = 6.3 m/s, we can easily find `dy.

`dy = v0^2 / (2 g)

which is easily evaluated to obtain `dy = 1.43 m.

MORE DETAILED 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).

STUDENT QUESTION

do we get dy'=v0^2/2g will this always be the case?

INSTRUCTOR RESPONSE

Most basic idea:

On the simplest level, this is a conversion of PE to KE. This is the first thing you should understand.

The initial KE will change to PE, so the change in PE is equal to the initial KE.

In this case the change in PE is m g `dy. For other situations and other conservative forces the expression for `dPE will be very different.

The simplest equation for this problem is therefore

init KE = increase in PE so that

1/2 m v0^2 = m g `dy

More general way of thinking about this problem:

More generally we want to think in terms of KE change and PE change. We avoid confusion by not worrying about whether each change is a loss or a gain.

Whenever conservative forces are absent, or being regarded as negligible, we can set the expression for KE change, plus the expression for PE change, equal to zero.

• In the present example, KE change is (final KE - initial KE) = (0 - 1/2 m v^2) = -1/2 m v^2, while PE change is m g `dy.

• We get the equation

-1/2 m v0^2 + m g `d y = 0.

• This equation is easily rearranged to get our original equation 1/2 m v0^2 = m g `dy.

The very last step in setting up the problem should be to write out the expressions for KE and PE changes.

• The expression for PE change, for example, depends completely on the nature of the conservative force. For gravitational PE near the surface of the Earth, that expression is m g `dy. For gravitational PE where distance from the surface changes significantly the expression would be G M m / r1 - G M m / r2. For a spring it would be 1/2 k x2^2 = 1/2 k x1^2.

• The expression for KE change is 1/2 m vf^2 - 1/2 m v0^2; this is always the expression as long as mass doesn't change.

In this particular case the equation will read

• 1/2 m vf^2 - 1/2 m v0^2 + m g `dy = 0

If we let vf = 0, the previous equations follow.

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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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:

½(950N/m)(.150m)^2= 10.7Joules= PE

½(.3kg)vf^2= 10.7Joules

vf= sqrt((2*10.7Joules)/.3kg)= 8.45m/s

dy= PE/mg

dy= 10.7Joules/.3kg(9.8m/s^2)= 3.64m

confidence rating #$&*:

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Given Solution:

`a

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’m confused by the elastic PE, what exactly does it mean? Also, I got the sign wrong on PE, I thought that since it was compressed the PE would be positive.

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Self-critique rating:

@& Elastic PE is always positive, relative to the unstretched or uncompressed state.

So then the spring is released, it will lose elastic PE. Thus `dPE_elastic is negative.

To compress the spring you need to exert a force in the direction of motion. The same is true if you stretch the spring (or stretch a rubber band).

In the case of the rubber band, as you have seen in the Force vs. Displacement experiment, you have to add up the F_ave * `ds contributions. If F is given as a function of position, you can integrate the function between the two positions.

For an ideal spring, which at position x exerts tension or compressive force - k x, you need to exert force k x. To stretch or compress the spring from equilibrium to position x, you exert a max force k x (at position x) and a min force zero (at the equilibrium position). Your average force is therefore (k x + 0) / 2 = 1/2 k x, exerted through displacement x, which requires work 1/2 kx * x = 1/2 k x^2. (This can also be obtained by integrating 1/2 k x with respect to x, from 0 to x).

When released the spring exerts equal and opposite forces as x returns to the equilibrium position, doing work 1/2 k x^2. Thus the compressed or stretched spring has the potential, upon release, to do work 1/2 k x^2. We call this 'elastic potential energy'.

In the current problem the spring stores elastic potential energy. Upon release the spring does work on the ball, and the net result is that the ball's gravitational PE increases.

*@

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Question: `qgen phy problem A high jumper needs to be moving fast enough at the jump to lift her center of mass 2.1 m and cross the bar at a speed of .7 m/s. What minimum velocity does she require?

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Your solution:

PE= m(9.8m/s^2)(2.1m)=20.58m^2/s^2 * m

KE= 1/2 m (.7m/s)^2= .245m^2/s^2 * m

20.58m^2/s^2 *m + .245m^2/s^2 *m = 20.825m^2/s^2 *m

20.825m^2/s^2 * m= ½ mv0^2

20.825m^2/s^2= ½v0^2

41.65m^2/s^2= v0^2

6.45m/s= v0

confidence rating #$&*:

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Given Solution:

`aFORMAL SOLUTION:

Formally we have `dPE + `dKE = 0.

`dPE = M * g * `dy = M * 20.6 m^2 / sec^2, where M is the mass of the jumper and `dy is the 2.1 m change in altitude.

`dKE = .5 M vf^2 - .5 M v0^2, where vf is the .7 m/s final velocity and v0 is the unknown initial velocity.

So we have

M g `dy + .5 M vf^2 - .5 M v0^2 = 0.

Dividing through by M we have

g `dy + .5 vf^2 - .5 v0^2 = 0.

Solving for v0 we obtain

v0 = sqrt( 2 g `dy + vf^2) = sqrt( 2* 9.8 m/s^2 * 2.1 m + (.7 m/s)^2 ) = sqrt( 41.2 m^2/s^2 + .49 m^2 / s^2) = sqrt( 41.7 m^2 / s^2) = 6.5 m/s, approx..

LESS FORMAL, MORE INTUITIVE, EQUIVALENT SOLUTION:

The high jumper must have enough KE at the beginning to increase his PE through the 2.1 m height and to still have the KE of his .7 m/s speed.

The PE change is M * g * 2.1 m = M * 20.6 m^2 / sec^2, where M is the mass of the jumper

The KE at the top is .5 M v^2 = .5 M (.7 m/s)^2 = M * .245 m^2 / s^2, where M is the mass of the jumper.

Since the 20.6 M m^2 / s^2 increase in PE must come at the expense of the initial KE, and since after the PE increase there is still M * .245 m^2 / s^2 in KE, the initial KE must have been 20.6 M m^2 / s^2 + .245 M m^s / s^2 =20.8 M m^s / s^2, approx.

If initial KE is 20.8 M m^s / s^2, then .5 M v0^2 = 20.8 M m^s / s^2.

We divide both sices of this equation by the jumper's mass M to get

.5 v0^2 = 20.8 m^2 / s^2, so that

v0^2 = 41.6 m^2 / s^2 and

v0 = `sqrt(41.6 m^2 / s^2) = 6.5 m/s, appprox.

STUDENT QUESTION

I used the equation 'dy=v0^2 / (2g). Isn't that easier?

INSTRUCTOR RESPONSE

Good, but that equation only applies under certain conditions. Your solution didn't account for the final KE, which doesn't make a lot of difference but does make enough to decide the winner of a competitive match.

In general you don't want to carry an equation like 'dy=v0^2/(2g) around with you. If you carry that one around, there are about a hundred others that apply to different situations, and you'll overload very quickly. Among other things, that equation doesn't account for both initial and final KE. It applies only when the PE change is gravitational, only near the surface of the Earth, and only when the final KE is zero. Way too many special conditions to keep in mind, way too much to remember.

You want to start your reasoning from `dKE + `dPE + `dW_noncons_ON = 0.

We assume that nonconservative forces are negligible, so that `dW_noncons_ON is itself zero, giving us

`dPE + `dKE = 0.

For this situation `dPE = m g `dy, `dKE = KE_f - KE_0 = 1/2 m vf^2 - 1/2 m v0^2, and the equation becomes

m g `dy + 1/2 m v0^2 - 1/2 m vf^2 = 0.

In a nutshell, there are only three things you need in order to analyze similar situations:

• `dKE + `dPE + `dW_noncons = 0

• KE = 1/2 m v^2

• `dPE = m g `dy (in the vicinity of the Earth's surface)

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Self-critique (if necessary):

Ok

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Self-critique rating:

ok

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Question: `qquery Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?

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Your solution:

½ (400N/m)(.220m)^2= 9.68PE

Sin(37)= -.64J

PE= 2kg(9.8m/s)dy

9.68J - .64J= 9.04PE

vf= sqrt((2*9.04Joules)/2kg)= 3m/s

dy= 9.04J/(2kg * 9.8m/s^2)= .46m

confidence rating #$&*:

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Given Solution:

`a** The spring exerts a force of 400 N / m * .220 m = 88 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 88 N) / 2 = 44 N, so the work done in the compression is

`dW = Fave * `ds = 44 N * .220 m = 9.7 Joules, approx.

If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 9.7 Joules / (2 kg) ) = 3.2 m/s, approx..

No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 9.7 J and the KE is zero, at which point it will begin to slide back down the incline.

After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds.

Setting this expression equal to KE we obtain the equation

.6 m g `ds = KE,

which we solve for `ds to obtain

`ds = KE / (.6 m g) = 9.7 Joules / (.6 * 2 kg * 9.8 m/s^2) = .82 meters, approx. **

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Self-critique (if necessary):

I messed up everything, I’m having a hard time understanding all the different scenarios and figuring out what I need to do with each. It all kind of runs together. How was the KE = 5J? I see how the dy = .6ds and how you use that to solve for ds.

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Self-critique rating:

@&

(note that sin(37) is the sine of 37 radians; of course I know that you mean sin(37 deg))

Energy is not a vector quantity. It doesn't have a direction. So you would never multiply work or energy by the sine or cosine of an angle.

The displacement of the block up the incline is a vector quantity. If the block travels distance `ds up the incline, its vertical coordinate increases by dy = `ds * sin(37 deg), and its gravitational PE therefore increases by m g `dy.

That 5.0 Joules should have been 9.7 Joules; that's an editing error I've apparently been overlooking for some time. Your solution for v_f was correct.*@

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Question: `qquery univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ.

What is the skier's speed at the bottom of the slope?

After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going?

Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?

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Your solution:

PE= 62kg(9.8m/s^2)(65m)= 39494Joules

39494J - 10.5J= 28994J= KE

vf= sqrt((2*28994J)/62kg)= 30.6m/s

28994J/82m= 353.6N

353.6N - 160N - .2(82m)= 177.2N

177.2N * 82m= 14530.4J

vf= sqrt((2*14530J)/62kg)= 21.6m/s

confidence rating #$&*:

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Given Solution:

`a** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ.

Formally we have

`dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ.

The speed of the skier at this point will be

v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx.

Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be

`dWnoncons = 280 N * 82 m = 23 kJ, approx.,

and the skier's KE will be

KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx.

This implies a speed of

v = sqrt( 2 KE / m) = 12 m/s, approx.

To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have

`dW = Fave * `ds, so that

Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N.

This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**

STUDENT QUESTION

If the PE is = 20.6, then why is the initial KE= 20.6 so that we are adding the .245 to the initial KE of 20.6 to get

20.8, I thought that the KE was equal and opposite to the PE why would we not subtract here?

INSTRUCTOR RESPONSE

`dKE + `dPE = 0, provided there are no nonconservative forces acting on the system.

In such a case, PE goes up as the mass rises, so KE goes down. Another way of looking at it: All or part of the KE converts to PE.

The mass can only go as high as the initial KE permits. Once the initial KE is 'used up', no increase in PE is possible (recall the assumption that no nonconservative forces act during this phase of motion).

At maximum height the mass is still moving in the horizontal direction, so not all of the KE converts to PE.

In this case PE increases by 20.6 M m^2/s^2, .245 M m^2/s^2 of KE is still present at the highest point, so about 20.8 M m^2 / s^2 of KE must have been present initially.

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Self-critique (if necessary):

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#*&!

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