Query 10

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course phy 201

I would particularly wish for feedback on this I am very unsure about this one

010. `query 10

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Question: `qQuery introductory problem set 3 #'s 7-12

Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.

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Your solution:

Displacement * net force

I have no idea of another way

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Given Solution:

First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds.

Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **

STUDENT QUESTION:

I wasn’t sure what equation to use to find KE the second way. What does Kef stand for?

INSTRUCTOR RESPONSE:

In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to stand for initial and final kinetic energies.

STUDENT QUESTION:

Ok I know the other equation now but I still don’t really understand it. How come you multiply each velocity by 0.5? I don’t really understand the second equation KE = 1/2 m v^2

INSTRUCTOR RESPONSE

On one level, KE = 1/2 m v^2 is simply a formula you have to know.

It isn't hard to derive that formula, as you'll see soon, and the 1/2 arises naturally enough. A synopsis of the derivation:

If force F_net is applied to mass m through displacement `ds then:

a = F_net / m, and

vf^2 = v0^2 + 2 a `ds

It's not difficult to rearrange the result of these two equations to get F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

You'll see the details soon, but that's where the formula KE = 1/2 m v^2 comes from; the 1/2 or 0.5 is part of the solution.

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Self-critique (if necessary):

While I can see that equals out I don't see why in the first place though

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By the definitions of velocity and acceleration as rates of change we get the equations of uniformly accelerated motion (which can also be obtained by integrating constant acceleration a twice with respect to clock time).

Newton's Second Law says that F_net = m * a. This has been very well verified, with increasing levels of accuracy and precision, over three centuries.

If we substitute a = F_net / m, which follows from Newton's Second Law, into the fourth equation of motion vf^2 = v0^2 + 2 a `ds, we easily rearrange to get

F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

We defined F_net * `ds to be work done by the net force, 1/2 m v^2 to be kinetic energy, and the result is that

`dW_net = `dKE.

This is why.

With your outstanding mathematics background, this argument should not be difficult to understand.

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Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of Physics students are encouraged, if they wish, to answer the question). In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?

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Your solution:

Not sure

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Given Solution:

In a nutshell:

since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2

since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds

Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2.

Thus F_net `ds is proportional to the change in v^2.

More detail:

It's very important in physics to be able to think in terms of proportionality.

To say that y is proportional to x is to say that for some k, y = k x.

That is, y is a constant multiple of x.

To say that a * `ds is proportional to the change in v^2 is to say that

for some k, a * `ds = k * ( change in v^2)--i.e., that

a * `ds is a constant multiple of the change in v^2.

In terms of the equations of motion, we know that

vf^2 = v0^2 + 2 a `ds so

a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2.

So a `ds is a constant multiple (1/2) of the change in v^2.

Formally we have

a `ds = k ( change in v^2) for the specific k value k = 1/2.

Now since Fnet = m a, we conclude that

Fnet * `ds = m a * `ds

and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get

Fnet `ds = m * k * (change in v^2), for k = 1/2.

Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2

With this revised value of k we have

Fnet * `ds = k * (change in v^2), where now k has the value m / 2.

That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity.

STUDENT COMMENT: I am still a bit confused. Going through the entire process I see how these

values correlate but on my own I am not coming up with the correct solution. I am getting lost after we discover the a `ds

is a constant multiple of (1/2) the change in v^2. Is it that I should simply substitute the k into the equation? Or am I

missing something else?

INSTRUCTOR RESPONSE: The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2 is the constant number 1/2).

F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the constant mass m).

Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2.

The conclusion is the F_net `ds is proportional to the change in v^2.

(The proportionality constant between F_net `ds and change in v^2 is 1/2 m.)

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Self-critique (if necessary):

Like earlier I can see that it does equal but thats all I don't really see why

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Question: How do our experimental results confirm or cause us to reject this hypothesis?

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Your solution:

Even after reading the answer Im not sure what we are asked for

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Given Solution:

The explanation for this result:

On a ramp with fixed slope the acceleration is constant so

a `ds is simply proportional to `ds

specifically a `ds = k * `ds for k = a.

In the preceding question we saw why

a * `ds = k * (change in v^2), with k = 1/2.

In our experiment the object always accelerated from rest. So

the change in v^2 for each trial would be from 0 to vf^2.

the change would therefore be just

change in v^2 = vf^2 - 0^2 = vf^2.

Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear.

The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2

We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of vf^2 vs. `ds.

(we could of course use that slope with our proportionality to determine a, if desired)

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Self-critique (if necessary):

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Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s.

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Your solution:

56 km/hr , 15.6 m/s , 53.33 ft/s

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Given Solution:

`aWe need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds.

We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches.

1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft,

5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km.

Thus

35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr.

We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s.

The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.

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Question: Openstax problem 2.16: A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s

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Your solution:

"""" That is a statement I have no way to determine what it is asking I will have to look at the answer to find what is asked

4 and 2/7 acceleration

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Given Solution:

It seems unlikely that the significant figures in this problem are realistic. What are the chances that an animal's speed and the corresponding time interval are both measured to 3-significant-figure accuracy (which might be realistic, though with different parts of the animal's body at any instant having different forward speeds the velocity measurement would be very challenging), and both yield such round numbers (1 in 100 chance of each means 1 in 10 000 chance this would occur).

However if we accept the significant figures specified for this problem, the result can be obtained as follows:

Acceleration is rate of change of velocity with respect to clock time, so that

acceleration = (change in velocity) / (change in clock time) = 30.0 meters / (7.00 seconds) = 4.29 meters / second.

Note that 30 / 7 = 4.28571428571... , but with 3-significant-figure information we can only be confident of our 3-significant-figure rounding of this result.

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Question: `qGen phy and prin phy prob 2.16: car accelerates uniformly from rest to 95 km/h in 6.2 s; find acceleration

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Your solution:

You can divide 95 by 3600

.02638 km/s divide that by 6.5 to get 0.00426 km/s^2

"""" I used a calculator for this

@&

It's fine to use a calculator for arithmetic you can't reasonably be expected to do in your head.

*@

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Given Solution:

`a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s.

So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s.

Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 s) = 4.2 m/s^2.

Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is

-4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.

STUDENT QUESTION:

How did we know that the final velocity was 0?

INSTRUCTOR RESPONSE:

The final velocity was 0 because the car came to rest.

Summary of what we were given:

Initial velocity is 95 km/hr, or 26.3 m/s.

Final velocity is 0, since the car came to rest.

The velocity makes this change in a time interval of 6.2 seconds.

We can easily reason out the result using the definition of acceleration:

The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity) / (change in clock time)

The change in velocity from the initial 0 m/s to the final 26.3 m/s is 26.3 m/s, so

acceleration = change in velocity / change in clock time = 26.3 m/s / (6.2 s) = 4.2 m/s^2.

We could also have used the equations of uniformly accelerated motion, with vf = 26.3 m/s, v0 = 0 and `dt = 6.2 seconds. However in this case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt = (26.3 m/s - 0 m/s) / (6.2 s) = 4.2 m/s^2.)

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Self-critique (if necessary):

"""" Just saying in the answer you said initial velocity is 95 km while in th equestion its starts from reat and ends at 95 km/hr

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Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.

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Your solution:

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Given Solution:

If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that

x1(t) = 25 m/s * t - .05 m/s^2 * t^2.

At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is

x2(t) = 200 m + 15 m/s * t .

The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation

25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds).

Rearranging the equation we have

-.05 t^2 + 10 t - 200 = 0.

The quadratic formula tells us that solutions are

t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 )

Simplifying we get solutions t = 22.54 and t = 177.46.

At t = 22.54 seconds the trains will collide.

Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply.

GOOD STUDENT SOLUTION:

for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt)

-10 = -.('dt)

'dt = 100

so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision.

'ds = 15 m/s(100) + 200 m

'ds = 1700 m

'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m.

The trains collide. **

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Self-critique (if necessary):

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Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.

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Your solution:

confidence rating #$&*:

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Given Solution:

If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that

x1(t) = 25 m/s * t - .05 m/s^2 * t^2.

At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is

x2(t) = 200 m + 15 m/s * t .

The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation

25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds).

Rearranging the equation we have

-.05 t^2 + 10 t - 200 = 0.

The quadratic formula tells us that solutions are

t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 )

Simplifying we get solutions t = 22.54 and t = 177.46.

At t = 22.54 seconds the trains will collide.

Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply.

GOOD STUDENT SOLUTION:

for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt)

-10 = -.('dt)

'dt = 100

so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision.

'ds = 15 m/s(100) + 200 m

'ds = 1700 m

'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m.

The trains collide. **

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Self-critique (if necessary):

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