query162

course phy201

query 16.2 016. `query 16

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Question: `qClass 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?

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Your solution:

Not sure

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Given Solution:

`a**

A quick synopsis:

The object accelerates uniformly downward, so the distance it falls is proportional to the square of the time of fall. Thus the time of fall is proportional to the square root of the distance fallen.

The object's horizontal velocity is constant, so its horizontal distance is proportional to the time of fall.

So the horizontal distance is proportional to the square root of the distance it falls.

More details:

The distance of vertical fall, starting with vertical velocity 0, is

`dy = v0 `dt + .5 a `dt^2 = 0 `dt + .5 a `dt^2 = .5 a `dt^2,

so `dy is proportional to `dt^2.

Equivalently, therefore, `dt is proportional to sqrt(`dy).

The horizontal distance is

`dx = v_horiz * `dt

so `dx is proportional to `dt.

`dx is proportional to `dt, and `dt is proportional to sqrt(`dy), so `dx is proportional to sqrt(`dy).

STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:

When the

object is accelerating downward the distance it falls is equal to the square root of the time it takes to fall.

Right idea but that would be the square of the time, not the square root.

So the time of our fall is equal to proportional to, not equal to; again you have the right idea the square root of the distance the object fell. Our horizontal velocity is constant which will make our horizontal distance equal to the time of the fall. Because of this, the horizontal distance in equal to the square root of the distance that our object falls.

Good.

To summarize:

The vertical distance is proportional to the square of the time of fall, so the time of fall is proportional to the square root of the distance fallen.

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Self-critique (if necessary):

Ok, so

dy or (ds) = v0'dt +.5a 'dt^2 which is the vertical, V0=0 vertically so ds= .5a'dt

therefore dy or ds is proportional to 'dt^2 , so 'dt is proportional to sqrt dy.

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Question: `qIn the preceding situation why do we expect that the vertical kinetic energy of the ball will be proportional to `dy?

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Your solution:

Vertical KE is proportional to dy because 'dw= fnet * 'ds and 'dw = KE. Work done on the ball as it drops vertically are equal to the kinetic energy working on the ball as it drops.

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Given Solution:

`a**

This could be argued by analyzing the motion of the object, and using the definition of kinetic energy:

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. **

In terms of energy the argument is simpler:

PE loss is -m g `dy.

Since m and g are constant for this situation, PE loss is therefore proportional to `dy. (This means, for example, that if `dy is doubled then PE loss is doubled; if `dy is halved then PE loss is halved.)

KE gain is equal to the PE loss, so KE gain is also proportional to `dy.

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Self-critique (if necessary):

Ok,

so we have the vf^2= v0^2 + 2a 'ds. To find the vertical velocity which shows that vertical velocity is proportional to 'ds fallen. Once we have the vf we can use the KE= .5mv^2 which is proportional to the sqrtVf as well as sqrt Of dy (ds). proving that KE is proportional to dy (or ds vertically).

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Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball?

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Your solution:

I am not sure hree, I thought that KE is equal to PE loss, so when KE is positive Pe would be negative but opposite to the vlaue of KE.

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Given Solution:

`a** There is some air friction, which dissipates some of the energy. PE is lost and the lost PE goes into an increase in KE, and into dissipated energy. The KE increase and dissipated energy 'share' the 'lost' PE.

STUDENT RESPONSE: Because actually some of the energy will be dissipated in the rotation of the ball as it drops?

INSTRUCTOR COMMENT: Good try. However there is KE in the rotation, so rotation accounts for some of the KE but doesn't dissipate 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. Dissipated energy is not recoverable.

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). **

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Self-critique (if necessary):

Ok, so air friction, dissipating energy, lowers the PE and Increases the KE.

??????????????????????So does this mean that they do not neccessarily have to be equal and opposite of each other??????????????????????

Good questions.

Check out this explanation, which was added to the original document:

Air friction doesn't lower the PE; it's

the change in altitude that lowers the PE (an object's gravitational potential

energy gets lower as it descends). The loss of PE results in an increase

in KE, though not as great an increase as if air friction wasn't present.

Related to the idea that `dKE and `dPE

should be equal and opposite (which is sometimes the case and sometimes not):

The energy situation is governed by
the work-energy theorem in the form

`dW_noncons_ON = `dKE + `dPE.

In general, a nonconservative force
can increase the KE and/or the PE in any way at all. It can increase one
without changing the other. `dKE and `dPE are not generally equal and
opposite. For example you can speed up a cart by pushing or pulling it along
a level surface. The force you exert is nonconservative, and it increases
the KE of the cart without changing its PE. Or you could lift an
object at a constant speed; its KE wouldn't change but its gravitational PE
would.

However in some situations nonconservative forces are either absent or
negligible. For example if you toss a steel ball a couple of meters into the
air and let it fall to the ground, it doesn't attain enough speed for air
resistance to become significant, so once you release it the ball pretty
much behaves as if nonconservative forces were absent. Then as it rises it
slows down, decreasing its KE and increasing its PE. As it then falls its PE
decreases but it speeds up, increasing its KE. Changes in KE and PE turn
out, in this case, to be equal and opposite.

Formally, since `dW_noncons_ON = `dKE
+ `dPE, it follows that if there are no nonconservative forces `dKE
+ `dPE = 0 and `dKE and `dPE are equal and opposite.

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Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr

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Your solution:
m=1250kg
v0=105km/hr
KE= .5m V^2
105km * 1000m/1km * 1 hr/3600s = 29.17m/s
KE= .5(1250kg) * (29.17m/s)^2
KE= 625kg * 850.89M/s^2
KE= 531,806.25J
'dw= 531,806.25J

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Given Solution:
`aThe 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.

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Self-critique (if necessary):
?????????????????????ok so the KE can be negative and the 'dw can be positive???????????????? Since the car is coming to rest I see why the KE would be negative, but I would think that the 'dw would also be negative ?????????????????

Another good question. My response is given below:

The given solution was more intuitive, and
wasn't completely specific in its application of the work-energy theorem.
It would have been more precise to say that to decrease the car's KE by 530 000
J, the car had to do 530 000 J of work against friction.

To be really consistent, careful use
of the _ON and _BY subscripts will be helpful:

In terms of the work-energy theorem:

`dW_net_ON = `dKE.

Since the car's KE decreases, `dKE is
negative and the net force acting on the car does negative work. We
conclude that

`dW_net_ON = -530 000 J.

The net force acting on the car is
friction, so it would be accurate to say that the frictional force acting on the
car does -530 000 J of work.

The car exerts an equal and opposite
frictional force on the road. So the work done BY the car is equal and
opposite to the work done ON the car:

`dW_net_BY = -`dW_net_ON = +530 000 J.

So the car does 530 000 J of
work, thus decreasing its KE by 530 000 J.

The problem as stated wasn't specific
about whether the work it requested was done on or by the car, so as it turns
out either answer could be considered correct.

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Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.

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Your solution:
I am not sure where to start with this one I think it hs something to do with KE= .5m V^2
but I am not sure
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Given Solution:
`aThe 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.

Brief summary of elastic PE, leaving out a few technicalities:

1/2 k x is the average force, x is the displacement so the work is 1/2 k x * x = 1/2 k x^2
F = -k x
Work to stretch = ave stretching force * distance of stretch
ave force is average of initial and final force (since force is linear)
applying these two ideas the work to stretch from equilibrium to position x is 1/2 k x * x, representing ave force * distance
the force is conservative, so this is the elastic PE at position x
STUDENT QUESTION:
What does the kx stand for?
INSTRUCTOR RESPONSE:
The premise is that when the end of the spring is displaced from its equilibrium position by displacement x, it will exert a force F = - k x back toward the equilibrium point.
Since the force is directed back toward the equilibrium point, it tends to 'restore' the end of the spring to its equilibrium position. Thus F in this case is called the 'restoring force'.
The force is F = - k x, with F being proportional to x, i.e,. the first power of the displacement. A graph of F vs. x would therefore be a straight line, and the restoring force is therefore said to be linear. A function is linear if its graph is a straight line.
So we say that F = - k x represents a linear restoring force.
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Self-critique (if necessary):
ok, so x= sqrt(2PE/k)
x= sqrt (2(25J)/440N/M)
x= sqrt(50J/440N/m)
x= sqrt.114
x= .34m
?????????????????Is this another formula, or is it a manipulation of a formula that we are already given????????
I am confused on what this means, I see that we end up with distance....I see how it is worked out, but I do not know what it means?????????????

Good questions. See my response:

The force exerted by a spring is F = - k
x, where x is how far it is stretched (positive x) or compressed (negative x):

If a spring is stretched beyond its
equilibrium length it 'pulls back' with a tension that increases the more it
is stretched. If x is how far it is stretched past its equilibrium length,
then a positive stretch x results in a tension force in the negative
direction.

For an ideal spring, a graph of force vs. stretch is a straight line through
the origin. The equation of this line is F = - k x. For any ideal spring k
is a constant number which, when multiplied by the stretch, then by -1
(because the direction of the tension is opposite that of the stretch),
gives us the tension force.

A negative value of x corresponds to compressing the spring rather than
stretching it. If a spring is compressed it 'pushes back', exerting a force
in the positive direction. For an ideal spring the equation F = - k x
continues to apply, since a negative value of x will appropriately result in
a positive value of F.

The constant number k is called
the spring constant.

At position x the potential energy of the
spring is 1/2 k x^2:

When stretched/compressed x units from
the equilibrium length, for the reasons indicated in the given solution, the
average force required is 1/2 k x and the displacement from equilibrium is x
so the work done is 1/2 k x * x = 1/2 k x^2.

The force exerted by an ideal spring is conservative, so between equilibrium
and position x the PE of the spring increases by amount 1/2 k x^2.

Taking the PE of the spring to be 0 when x = 0, it follows that at position
x its PE is 1/2 k x^2:

PE_spring = 1/2 k x^2.

In the current example the spring constant
is k = 440 Newtons / meter, and we want to find the stretch needed to give us PE
of 25 Joules.

We therefore solve the equation

PE = 1/2 k x^2

for x, obtaining

x = +- sqrt( 2 PE / k),

and substitute our given values of k and PE to get the result.

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Question: `qgen 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?

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Your solution:
m= 88g 88g * 1kg/1000g= .088kg
'ds= 78cm or .78m
fave= 110N
'dw= fnet * 'ds
KE= .5m v^2
'dw = 110N * .78m
'dw= 85.8J
KE= 85.8J
85.8J= .5(.088) v^2
85.80= .044 v^2
v^2= 1950
v=44.16m/s

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Given Solution:
`a** 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.. **

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Self-critique (if necessary):
ok, I understand

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Question: `qquery 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?

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Your solution:

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Given Solution:
`a** 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) = 89 m/s, approx. **

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Self-critique (if necessary):

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Question: `qUniv. 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?

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Your solution:

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Given Solution:
`a** 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. **

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Self-critique (if necessary):

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&#Good responses. See my notes and let me know if you have questions. &#

query162

Good questions. Check out this explanation, which was added to the original document:

Another good question. My response is given below:

Good questions. See my response:

&#Good responses. See my notes and let me know if you have questions. &#

Good questions.

Check out this explanation, which was added to the original document: