Query 172

course PHY201

17.2017. `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:

V0= 5.3m/s

a= 9.8m/s^2

Vf = v0 + a * 'dt

5.3m/s = 0 + 9.8m/s^2 * 'dt

'dt= .54s

ds= vave * 'dt

ds= 2.65m/s * .54s

'ds= 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 / 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).

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Self-critique (if necessary):

ok so using the KE= 1/2 m v^2

divide out the m, or divide it by both sides to get g * 'dy = 1/2 vo^2

so 9.8m/s^2= 1/2 (5.3m/s)^2

= 1(5.3m/s)^2/(2* 9.8m/s^2)

= 1.43m

I used the uniform acceleration to solve this it seemed easier, but Ithink i am starting to understand this one

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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:

950n/m * .150m = 142.5n

F= m * a

142.5n = .30kg * a

a= 475m/s^2

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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):

OK

Pe stored in the spring is .5k x^2

= .5(950N/m) (.150m)^2= 10.7J

Gravitational PE= m g 'dy

= .30kg * 9.8m/s^2 * .150m = .44J

So total change in PE = -10.7 + 4.4J= -10.3J

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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= .5k x^2

PE= m g 'dy

we know

'dy=2.1m

PE= m 9.8m/s^2 * 2.1m

PE= 20.58 Mm^2/s^2

Pe= .5(20.58m^2/s^2) (.7m/s)^2

PE=5.04

KE= -5.04

-5.04= 1/2m v^2

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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.

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Self-critique (if necessary):

KE= .5m v^2

=.5m (.7m/s)^2

.5m .49m^2/s^2

=.245M m^2/s^2

KE= .245M m^2/s^2 + 20.6M m^2/s^2 = 20.8M m^2/s^2

so

20.8M m^2/s^2= .5m vf^2

vf^2=20.8M m^2/s^2 /.5M

vf^2= 41.6 m^2/s^2

vf=6.449m/s

Ok, I think I am starting to get this, the M does not factor out until you use the equation for KE, so we find the PE first witht he

equation PE= m g * dy; m= mass g= gravity and dy= vertical displacement, We then use the information to find the KE, using KE= .5mvf^2,

which was .245 and add that to the Initial KE

????????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?

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

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