Asst 14 Query

course Phy 121

×ÖêÂþÉ¤à¤—ßJëpÅõÔ}–¥—‰úèÕŒúTÁ©assignment #014

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014. `query 14

Physics I

06-24-2006

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00:29:11

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 -->

When we multiply the acceleration by the time interval and initial velocity is zero, we get the final velocity.

Starts from rest: vo = 0 m/s

a = `dv/`dt = (vf - vo)/`dt = vf/`dt

a = vf/`dt

vf = a * `dt

We can find the distance traveled by using one of the four equations of uniformly accelerated motion:

`ds = (vo)(`dt) + 1/2(a)(`dt^2)

Substitue Fnet/m for acceleration:

`ds = (vo)(`dt) + 1/2(Fnet/m)(`dt^2)

Initial velocity is zero:

`ds = 1/2(Fnet/m)(`dt^2)

When we multiply net force by the displacement, we get

Fnet * `ds = (Fnet)[1/2(Fnet/m)(`dt^2)]

When we distribute Fnet:

Fnet * `ds = 1/2(Fnet^2/m)(`dt^2)

Kinetic energy is equal to the work done on the object which is the net force multiplied by the displacement.

KE = `dWnet = Fnet * `ds = 1/2(Fnet^2/m)(`dt^2)

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00:31:54

**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 -->

I understand this problem.

Which of these explanations is the best way to solve for displacement for this problem? I didn't use average velocity.

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00:32:26

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 -->

Kinetic energy of the system when work is done by the system against nonconservative forces is not conserved. It is spent and the kinetic energy decreases.

Kinetic energy of the system when work is done by the system against conservative forces is conserved. The kinetic energy will increase or simply be conserved.

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00:34:03

** 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.

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00:35:18

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 amount of work done to stretch the rubber band in order to pull the rail back is equal to the amount of work done by the rubber band on the rail.

The work done by the rubber band on the rail when the rail is released is equal to the work done by the rail against friction. We know that this must be true, or the rail would not be able to overcome the force of friction and would not move.

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00:36:15

** 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 -->

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00:37:46

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

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RESPONSE -->

For the rail:

Since F = m * a, when we multiply by the displacement, F * `ds = (m*a)`ds

We know from one of the equations of motion that vf^2 = vo^2 + 2(a)(`ds) which can be rearranged to read:

a * `ds = 1/2(vf^2 - vo^2)

From these two equations we can see that force and displacement are proportional to the change in velocity.

The rail is moved by the force of the rubber band. Therefore, the rubber band is causing the change in velocity of the rail. From this we can see that the distance traveled by the rail is proportional to the force of the rubber band. What this means is the force of the rubber band and the distance stretched is equal to the force of friction holding the rail back and the distance the rail moves.

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00:38:57

** 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 understand this concept. Did I misinterpret anything about the forces or distance?

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00:39:12

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 -->

I am not required to answer this question.

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00:39:16

** 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.

(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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00:39:35

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 -->

I am not required to answer this problem.

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00:40:26

** If a landing craft slows then its acceleration is in the direction opposite to its motion, therefore upward.

If it speeds up while landing that its acceleration is in the direction of its motion, therefore downward.

If the upward motion is taken as the positive direction, then the acceleration under a thrust of 25 kN is + 1.2 m/s^2, and the acceleration when under thrust of 10 kN is - .8 m/s^2.

In either case m * a = net force.

Net force is thrust force + gravitational force.

1 st case, net force is 25 kN so m * 1.2 m/s/s + m * g = 25 kN.

1 st case, net force is 10 kN so m * (-.8 m/s/s ) + m * g = 10 kN.

Solve these equations simultaneously to get the weight m * g (multiply 1 st eqn by 2 and 2d by 3 and add equations to eliminate the first term on the left-hand side of each equation; solve for m * g).

The solution is m * g = 16,000 kN.

Another solution:

In both cases F / a = m so if upward is positive and weight is wt we have

(25 kN - wt) / (1.2 m/s^2) = m and

(10 kN - wt) / (-.8 m/s^2) = m so

(25 kN - wt) / (1.2 m/s^2) = (10 kN - wt) / (-.8 m/s^2).

Solving for wt we get 16 kN. **

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RESPONSE -->

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Good. Let me know if you have questions.

I've been including the following explanation, related to question 8 on this query, to most students after they get to this point. You don't really appear to need this, but you might find it useful so here it is for your reference:

The KE of an object is always 1/2 m v^2, where m is its mass and v its velocity.

The KE of an object, and therefore its velocity, is affected by the net force acting on the object, which is equal and opposite to the net force exerted by the object. In every situation you have to know whether you are thinking from the perspective of the object (also called 'the system') or the forces acting on the object (or system). In some situations, it is most natural to think from one point of view or the other, and depending on the situation it could be either perspective. In some situations, it's easy to confuse the two perspectives, to think of some aspects from one point of view and some aspects from the other. This is the key to much of the confusion that arises when thinking about energy.

The first thing to understand in the present situation, and I believe you understand it, is that the automobile starts with 10,000 Joules of energy, and it going up the hill it's going to gradually lose that energy until it's all gone, at which point the automobile will stop moving.

Depending on the situation, the automobile might or might not then start moving back down the hill (in this case it would do so), and would start building KE. If there is no frictional or other dissipative loss, by the time it gets back to its original altitude its KE will have built back to its original value.

If friction is involved, it will lose energy to friction going up, so it won't go as far up the hill, won't have as far to coast to get back to its starting point, and will lose additional energy to friction on the way back down.

The work done by a force F is F * `ds, where both F and `ds are each positive or negative, depending whether each is in the direction you have chosen to be positive, or opposite to this direction. F * `ds is therefore either a positive or negative quantity.

If a net force F_net_ON is acting ON the system, then the work done by that net force will change the KE of the system accordingly, increasing the KE if it does positive work, decreasing the KE if it does negative work.

If a system itself exerts a net force F_net_BY AGAINST the forces acting on the system, then the work done BY the system is equal to the work done by the force F_net_BY, and if the system does positive work, it is at the expense of its KE.

The forces F_net_BY and F_net_ON in the preceding two paragraphs are equal and opposite, by Newton's Third Law.

All this can be summarized by the two equivalent formulas

`dW_net_ON = `dKE and

`dW_net_BY = - `dKE.

Soon `dW_net_ON and `dW_net_BY will be broken up into two parts, the part associated with conservative forces and the part associated with nonconservative forces. But for the present, there are really only a few basic formulas involved:

Newton's Second Law and its application to gravitational force:

F_net = m * a

F_grav = weight = m * g, where g is the acceleration of gravity

KE = 1/2 m v^2

`dW = F * `ds

and the two equivalent formulas

`dW_net_ON = `dKE and

`dW_net_BY = - `dKE.

Do you see how these formulas fit in to the given solution?

There are a few other ideas involved in situations involving motion up and down an incline, or the motion of a system of two weights over a pulley:

These systems typically involve motion which can be in one specific direction, or opposite that direction. You get to choose which of these two directions is positive.

On an incline the gravitational force is equivalent to two forces, one parallel to and one perpendicular to the incline. These forces are ultimately found using trigonometry, but at this point in the course we rely on the approximation that the component of gravitational force down an incline with small slope is very well approximated the product of the slope and the gravitational force (e.g., on a slope of .1 the component of the gravitational force down the incline on an object weighing, say, 50 Newtons is .1 * 50 Newtons = 5 Newtons).

We also speak of the frictional force as a certain percent of the weight of the system. We will later refine this idea, but it applies in the examples used here to a very good approximation. So for example if friction is said to be 3% of the weight of the system, a 50 Newton system would experience a frictional force of .03 * 50 N = 1.5 N. This force would be in the direction opposite that of the system's motion, and as long as the direction of motion happens to be the direction originally chosen as positive, the frictional force would be negative; when the direction of motion is negative the frictional force would be positive. Note that if the system changes its direction of motion, the frictional force also changes its direction of motion.