query assignment 14

course phy 201

ƒassignment #014

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014. `query 14

Physics I

10-12-2007

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13:23:48

set 3 intro prob sets

If you calculate the acceleration on a mass m which starts from rest under the influence of a constant net force Fnet and multiply by a time interval `dt what do you get?

How far does the object travel during this time and what velocity does it attain?

What do you get when you multiply the net force by the distance traveled?

What kinetic energy does the object attain?

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RESPONSE -->

Know: a vo=0 m Fnet dt

a * dt = dv or, in this case vf which is the velocity it attains.

You need to answer in terms of the given quantities. Fnet and m are given; a is not but is computed from Fnet and m.

So your answer should read

Fnet / m * `dt = `dv,

which is consistent with the given solution.

ds = vAVG (vf + v0)/2 * dt

Fnet * ds = W or dKE

This is not expressed in terms of m and Fnet. The question was phrased in terms of Fnet, m, `dt and `ds.

confidence assessment: 2

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13:27:44

**STUDENT ANSWER AND INSTRUCTOR COMMENTS: a*'dt = the final velocity if V0=0. to get the change in position you would divide the final velocity(since V0=0) by 2 to get the average velocity and then multiply that by the 'dt to get the units of distance traveled.

Multiply that by the 'dt to get the units of distance traveled. It attains a Vf of a*'dt as shown above because V0=0, if V0 was not zero you would have to add that to the a*'dt to get the final velocity.

When you multiply Fnet by 'dt you get the same thing you would get if you multiply the mass by the change in velocity(which in this case is the same as the final velocity). This is the change in momentum. The Kinetic Energy Attained is the forcenet multiplied by the change in time.

a = Fnet / m. So a `dt = Fnet / m * `dt = vf.

The object travels distance `ds = v0 `dt + .5 a `dt^2 = .5 Fnet / m * `dt^2.

When we multiply Fnet * `ds you get Fnet * ( .5 Fnet / m * `dt^2) = .5 Fnet^2 `dt^2 / m.

The KE attained is .5 m vf^2 = .5 m * ( Fnet / m * `dt)^2 = .5 Fnet^2 / m * `dt^2.

Fnet * `ds is equal to the KE attained.

The expression for the average velocity would be [ (v0 + a * `dt) + v0 ] / 2 = v0 + 1/2 a `dt so the displacement would be (v0 + 1/2 a `dt) * `dt = v0 `dt + 1/2 a `dt^2. This is equal to (v0 `dt + 1/2 a `dt^2) * Fnet = (v0 `dt + 1/2 a `dt^2) * m a , since Fnet = m a. **

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RESPONSE -->

Not sure if I got this right or not. The approach taken in the answer may be different then mine.

KE = Fnet * dt??? I thought it was Fnet *ds

It is. The given solution never relates Fnet * `dt to KE.

self critique assessment: 1

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13:31:11

Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system. How does PE come into this relationship?

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RESPONSE -->

dKE is the work done ON the system by the Fnet

dPE is the work done BY the system against conservative forces

confidence assessment: 2

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13:32:34

** The work done by the system against all forces will decrease the KE by an equal amount. If some of the forces are conservative, then work done against them increases the PE and if PE later decreases this work will be recovered. Work done against non-conservative forces is not stored and cannot be recovered.

STUDENT RESPONSE WITH INSTRUCTOR COMMENTARY: The work done by a system against nonconservative forces is the work done to overcome friction in a system- which means energy is dissipated in the form of thermal energy into the 'atmosphere.'

Good. Friction is a nonconservative force.

However there are other nonconservative forces--e.g., you could be exerting a force on the system using your muscles, and that force could be helping or hindering the system. A rocket engine would also be exerting a nonconservative force, as would just about any engine. These forces would be nonconservative since once the work is done it can't be recovered.

STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The work done by a system against conservative forces is like the work to overcome the mass being pulled by gravity.

INSTRUCTOR COMMENT: not bad; more generally work done against conservative force is work that is conserved and can later be recovered in the form of mechanical energy **

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RESPONSE -->

ok

self critique assessment: 3

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13:38:10

class notes: rubber band and rail

How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest?

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RESPONSE -->

The rubber band is a conservative force (minus heat losses). So as you pull the rubber band you are doing work on it and increasing its PE. I think the arm doing the pulling would experience an equal loss in KE.

When the rubber band propels the rail it releases and reduces its PE and converts it into KE in the rail although some of the energy is lost as friction and does not add to the KE of the rail.

As the rail fights friction it loses KE and so does negative work against the friction?

confidence assessment: 2

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13:39:39

** The work done to stretch the rubber band would in an ideal situation be available when the rubber band is released.

Assuming that the only forces acting on the rail are friction and the force exerted by the rubber band, the work done by the rail against friction, up through the instant the rail stops, will equal the work done by the rubber band on the rail.

Note that in reality there is some heating and cooling of the rubber band, so some of the energy gets lost and the rubber band ends up doing less work on the rail than the work required to stretch it. **

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RESPONSE -->

I understand. The KE imparted by the rubber band on the rail will eventually equal the work done by the rail against friction.

self critique assessment: 3

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13:42:28

Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?

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RESPONSE -->

The total W represented by the rubber band releasing is F*ds of the rubber band and assuming that friction is constant, the distance the rail will travel before stopping will be proportional to the amount of W (energy) imparted by the rubber band.

confidence assessment: 2

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13:43:47

** The F `ds total of the rail when it is accelerated by the rubber band is equal Fave `ds, which is equal to to m * aAve * `ds. Here aAve is the average acceleration of the rail by the rubber band.

2 aAve `ds = vf^2 - v0^2 by the fourth equation of motion. So the F `ds total is proportional to the change in v^2.

The rail is then stopped by the frictional force f; since f `ds is equal to m * a * `ds, where a is the acceleration of the sliding rail, it follows that f `ds is also proportional to the change in v^2.

Change in v^2 under the influence of the rubber band (rest to max vel) is equal and opposite to the change in v^2 while sliding against friction (max vel back to rest), so work f `ds done by friction must be equal and opposite to F `ds.

This ignores the small work done by friction while the rubber band is accelerating the rail. **

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RESPONSE -->

I think this says the same thing in a different way.

self critique assessment: 2

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14:04:17

gen phy A person of mass 66 kg crouches then jumps to a height of .8 meters. From the crouches position to the point where the person leaves the ground the distance is 20 cm. What average force is exerted over this 20-cm distance?

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RESPONSE -->

known: m = 66kg ds of jump=.8m ds prior to jump = 20 cm

Want to know: Favg for first 20cm of ds

Can't get this one.

confidence assessment: 1

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14:09:14

** the normal force is the force between and perpendicular to the two surfaces in contact, which would be 646.8N if the jumper was in equilibrium. However during the jump this is not the case, and the normal force must be part of a net force that accelerates the jumper upward.

In a nutshell the net force must do enough work to raise the person's weight 1 meter while acting through only a .2 meter displacement, and must therefore be 5 times the person's weight. The person still has to support his weight so the normal force must be 6 times the person's weight.

The detailed reasoning is as follows:

To solve this problem you have to see that the average net force on the jumper while moving through the `dy = 20 cm vertical displacement is equal to the sum of the (upward) average normal force and the (downward) gravitational force:

Fnet = Fnormal - m g.

This net force does work sufficient to increase the jumper's potential energy as he or she rises 1 meter (from the .20 m crouch to the .8 m height). So

Fnet * `dy = PE increase,

giving us

( Fnormal - m g ) * `dy = PE increase.

PE increase is 66 kg * 9.8 m/s^2 * 1 meter = 650 Joules approx.

m g = 66 kg * 9.8 m/s^2 = 650 Newtons, approx..

As noted before `dy = 20 cm = .2 meters.

So

(Fnormal - 650 N) * .2 meters = 650 Joules

Fnormal - 650 N = 650 J / (.2 m)

Fnormal = 650 J / (.2 m) + 650 N = 3250 N + 650 N = 3900 N.

An average force of 3900 N is required to make this jump from the given crouch.

This is equivalent to the force exerted by a 250-lb weightlifter doing a 'squat' exercise with about 600 pounds on his shoulders. It is extremely unlikely that anyone could exert this much force without the additional weight.

A 20-cm crouch is only about 8 inches and vertical jumps are typically done with considerably more crouch than this. With a 40-cm crouch such a jump would require only half this total force and is probably feasible. **

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RESPONSE -->

I'm not surprised I couldn't get this. Just a little too advanced for me at this stage when my understanding of force and energy are not too good.

self critique assessment: 2

In a smaller nutshell:

The work done while the player is in contact with the floor (where `ds = 20 cm) must be sufficient to increase his PE from his starting position to the top of his jump (1 meter above his crouched position). Easy to find change in PE, which tells you how much work was required; then since you know the displacement through which the work was done you easily find the average force.

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14:10:09

univ phy text prob 4.42 (4.40 in 10th edition) Mercury lander near surface upward thrust 25 kN slows at rate 1.2 m/s^2; 10 kN speeds up at .8 m/s^2; what is weight at surface?

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RESPONSE -->

n/a

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

confidence assessment: 3

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