Physics Question + Answer

[The Law]

Think it'd be useful if we had a physics Q + A thread. I can't be the only one who has physics questions (or maybe I am ).

[stefanci]

ask one! let's see what happens!

[The Law]

Hmm, I'm struggling with this EK question... so if anybody can try to explain it, would be awesome.

 

A man puts springs on the bottoms of his shoes and jumps off a 10 m cliff. Assuming the springs follow Hooke's law, from the moment the springs touch the ground to when they are fully compressed, the magnitude of the man's acceleration will:

 

a) decrease then increase

increase then decrease

c) remain constant at 10 m/s^2

d) remains constant at zero

 

Highlight Line for Answer: _A_

[HBP]

Hmm, I'm struggling with this EK question... so if anybody can try to explain it, would be awesome.

 

The magnitude has to decrease to 0 for the man to move at a constant speed, but then the magntitude must increase (opposite direction) for the man to stop.

 

-10 --> 0 --> +whatever

 

 

Although the acceleration is always becoming more positive, the magnitude decreases then increases.

[Avinash]

Hmm, I'm struggling with this EK question... so if anybody can try to explain it, would be awesome.

 

I hope this makes sense, if not, some1 feel free to add a less convoluted explanation.

 

Fnet=ma

When the man is falling, his Fnet = force of gravity = mg = ma.

As soon as he hits the ground Fnet becomes Force of Gravity + Force Spring.

The spring will be exerting a force of kx, where k is the spring constant and x is the distance of compression. Also, remember, the spring will be exerting a force opposite to the direction of gravity, in other words -kx.

 

So, once he touches the ground,

Fnet= Force of Gravity + (-Force of Spring) = mg + (-kx) = ma.

 

As the spring is compressed more (x gets larger), Fnet get's smaller, because mg is constant, and we're subtracting a larger number from it. Since Fnet is getting smaller, this must mean that acceleration is decreasing.

 

Once x reaches a value, such that kx = mg, then Fnet=0 and a=0. Once x gets larger, such that kx > mg, then Fnet will become negative, meaning the "magnitude" of his acceleration will become a whole number again, but in the opposite direction.

 

So that must mean his acceleration starts off at 10m/s^2, decreases to 0m/s^2 and then increases again, in the opposite direction.

[dan0105]

man tricky question, it really isn't obvious to just think about.

 

But ummm, here it goes,

right when the springs come in contact with the ground we can do a force diagram for the guy's feet.

 

F = mg (down from weight) - kx (up from springs)

thus the accel is, F = ma = mg - kx, or,

a = (mg-kx)/m

 

initially, k is super small and is almost unnoticable, but it still makes the acceleration less than the 9.8m/ss

 

(mg-k(0.00001))/m slightly less than (mg-k(0))/m = mg/m = g.

 

After some more falling, the x gets bigger because the person kept moving down, and this makes the spring force stronger, so we get something like

 

(mg-k(2))/m which is now a LOT smaller than g, but at a certain point (the maximum compression, and thus maximum x value), the kx term will become a lot bigger than the constant mg term, and so we will get the accel to be big again, but in the opposite direction (now the spring direction)...

[dan0105]

holy crap, in the time it took me to write that, two other people beat me to it, how many people are on tonight!!!

[The Law]

Ooh, I see it! That question was really tricky, but now it doesn't look so bad. Thanks everyone!

[gamma]

if in a parallel plate capacitor, you take electrons from the + plate and move them to the - plate, and you have to determine the amount of work done, why do you have to make the positive potential difference --> negative potential difference ?

[stefanci]

if in a parallel plate capacitor, you take electrons from the + plate and move them to the - plate, and you have to determine the amount of work done, why do you have to make the positive potential difference --> negative potential difference ?

 

You're creating potential energy, thereby you're causing work on the system, which is positive. Since W = F*dcostheta, you're causing a force on the electron (since it wants to repel from the negative side), so the work is positive, and so its the potential difference.

 

U = q*deltaV

 

....from my understanding.

[tallguy81]

New Question!

 

For Waves:

Amplitude = Loudness (obviously)

Frequency = Pitch

__________ = Tone

^

^

^

Fill in the blank plz!

[Shadowplay]

New Question!

 

For Waves:

Amplitude = Loudness (obviously)

Frequency = Pitch

__________ = Tone

^

^

^

Fill in the blank plz!

 

I want to say wavelength but I'm not sure - basing the answer entirely on a physics lab involving pipes open at one end only: different numbers of half wavelength in the pipe changed the tone we heard...

[dan0105]

Can't totally guarantee it, but I'm pretty sure "tone" just means a sound at certain frequency which ALSO has the "proper" and repeating sin wave (sinusoidal curve). Like there a lot of sounds that can be made at a certain frequency but they don't necessarily have to have the sin wave pattern like the open and closed ended pipes do, tone on the other hand implies this.

[tallguy81]

Can't totally guarantee it, but I'm pretty sure "tone" just means a sound at certain frequency which ALSO has the "proper" and repeating sin wave (sinusoidal curve). Like there a lot of sounds that can be made at a certain frequency but they don't necessarily have to have the sin wave pattern like the open and closed ended pipes do, tone on the other hand implies this.

 

Thanks for your answer. But should I apply a "tone" to the wave's fundamental frequency or also its following harmonics (depending upon the closed/open pipe system associated).

[AndyDude]

Plane is inclined at 30 deg angle. Coefficient of kinetic friction 0.1. Mass 100kg. Mass is moving up the inclined planed at velocity of 2m/s. What Tension should be applied to the rope in order to make the mass reverse direction in exactly 1s?

 

a) 213 387 c) 613 d) 787

 

Courtesy: Examkrackers © 2003.

[Avinash]

Is the rope pulling the block up the plane, or is the rope pulling it down?

[AndyDude]

Mass is moving up, rope is pulling up on the mass on the inclined plane.

[AndyDude]

Which of the following represents maximum height reached by a projectile?

 

a) v^2 / 2g

1/8 gt^2

c) vt

d) vt sin ø

[Avinash]

Plane is inclined at 30 deg angle. Coefficient of kinetic friction 0.1. Mass 100kg. Mass is moving up the inclined planed at velocity of 2m/s. What Tension should be applied to the rope in order to make the mass reverse direction in exactly 1s?

 

a) 213 387 c) 613 d) 787

 

Courtesy: Examkrackers © 2003.

 

For convention, down will be negative, up will be positive.

 

For the block to completely reverse direction, it must reduce it's velocity to 0m/s.

The problem requires it to reverse direction in 1sec, so that must mean the acceleration must be -2m/s^2.

 

Vf=Vi+at

0=2 + a(1)

-2=a

 

To give the block a -2m/s^2 acceleration, the net force on the block must be:

Fnet = ma

Fnet = (100)(-2)

Fnet = -200N (200N in the downwards direction)

 

The forces pulling the block in the down (negative direction) are mgsin30 and mgcos30*co.kinetic.fric and the tension of the rope pulls it upwards.

 

Fnet = T + mgsin30 + mgcos30*co.kinetic.fric

-200 = T + (100)(-10)(0.5) + (100)(-10)(0.87)(0.1)

-200 + 500 + 87 = T

T=387N (since it's positive, it's upwards).

[Avinash]

Which of the following represents maximum height reached by a projectile?

 

a) v^2 / 2g

1/8 gt^2

c) vt

d) vt sin ø

 

Vf^2-Vi^2 =2ad

 

At the max height, the velocity is zero.

 

0 - Vi^2 =2ad

-Vi^2 = 2ad

-Vi^2 /2a = d

[AndyDude]

Avinash, your answer to the tension is correct.

 

For the max. height, I solved the question similar to yours. But that's not the right answer. The solution divides the projectile motion into half and half. Focusing on the latter half we know at max. height, v = 0m/s. This is our initial v. d = vt + 1/2 a*t^2 (given v= 0)

 

d = 1/2 a*t^2 ; since we are using only half of the total projectile motion, t = 1/2 the total time.

 

d = 1/2 a*(1/2t)^2 = 1/8 a*t^2. (a=g)

 

It's quite an interesting question. I guess your solution would be right, but it gets complicated if we aren't given the initial velocity.

[Sutler]

permeability of free space -- i know it's a constant and when it's used but wtf does it really mean?

[The Law]

This is based on a diagram but basically here is what you have:

 

Pretend you have a point charge, Q. A second point charge (with negligible electric field) q is located at a distance 2r. If q1 and q1 are both positively charged, and q is released, what is the max velocity that can be achieved by q2? Let m be the mass of q.

 

A) sq rt KQq/mr

 

sq rt kQq/mr^2

 

C) sq rt 2kQq/mr^2

 

d) sq rt 2kQq/mr

 

 

 

 

Answer (highlight to see): _D_

 

I did not get that answer, I got instead answer A. This is my reasoning:

 

Basically, I tried saying the initial potential energy is equal to the final kinetic energy (and thus the point of highest velocity of q). The initial potential energy is equal to KqQ/2r. The final kinetic energy is 1/2mv^2.

 

So we started off at a point 2r, and the PEi = KEf

 

1. kQq/2r = 1/2 mv^2

 

2. 2kQq/2mr = v^2

 

3. v = sq rt of KQq/mr

 

What am I doing wrong? It's driving me a tiny bit crazy.

[The Law]

Avinash, your answer to the tension is correct.

 

For the max. height, I solved the question similar to yours. But that's not the right answer. The solution divides the projectile motion into half and half. Focusing on the latter half we know at max. height, v = 0m/s. This is our initial v. d = vt + 1/2 a*t^2 (given v= 0)

 

d = 1/2 a*t^2 ; since we are using only half of the total projectile motion, t = 1/2 the total time.

 

d = 1/2 a*(1/2t)^2 = 1/8 a*t^2. (a=g)

 

It's quite an interesting question. I guess your solution would be right, but it gets complicated if we aren't given the initial velocity.

 

This always confuses me. When we start looking at only half of the projectile, I'm always worried the equations aren't working. I don't see why it wouldn't though, the acceleration is constant. Usually, I'm concerned when it comes to the "t" because if we look at half the projectile, then t must be halved. I would have solved it just like avinash did - there is no t in that case. I wonder if anyone could explain why that method is incorrect.

[srindogg@hotmail.com]

The initial potential energy is equal to KqQ/2r. The final kinetic energy is 1/2mv^2.

 

QUOTE]

 

Formula mistake: The initial potential energy is KqQ/r