A problem from college physics course :-(

[Andy636]

Originally Posted By: IndyIan
Originally Posted By: Andy636
Originally Posted By: IndyIan

Wouldn't 0.048m/s^2 be a reasonable coasting decelleration rate for a car at 20m/s?
I'm too lazy calculate how many hp that would be...


You would stop in 4.1666666 Km

You don't have a total vacuum lane on your highways over there?

What are these highways you are talking about?

 

[MolaKule]

Originally Posted By: Vikas
"A race car accelerates from 10 m/s to 50
m/s in a distance of 3 m. The mass of the
car is 2500 kg. If there is a force of
resistance of 120 N acting against the car,
find the average value for the force
propelling the car. "

Can we all agree that this professor probably bikes to the university and has no idea about how much a typical car (or race car) weighs? This is pretty sad state for a physics professor who is not in habit of noticing something odd about the numbers. By the way, this is one of the top ranked University in Boston :-( With these type of professors, how do we expect students to excel in real life?


It doesn't matter what he drives or the actual physical values in the problem, he wants to know if you understand the physical concepts and be able to solve the problem.

BTW, any resistance is a vector force opposing the force due to acceleration.

 

[paulo57509]

Originally Posted By: Vikas
.....Time to open the Resnick&Haliday and if see those two guys had any crazy exercises like this in that book!


I still have both volumes on my book shelf someplace.

 

[KevGuy]

Originally Posted By: Andy636
To be fairly honest I am a bit disappointed that most of the people took the time to notice how big the acceleration is but nobody noticed that the friction coefficient is 0.004892.

Bitog members are getting soft these days

That's assuming that the resistive force is due entirely to friction. It may account for wind resistance/ drag too.

 

[KevGuy]

This car with a 40 g acceleration must have an orange can on the engine, otherwise the engine would not be lubricated properly to do it.

 

[Vikas]

Continuing on this college physics course:-

I have no problems solving the typical centripetal force/acceleration problems such as roller coasters, stunt plane loops or car coasting over the top of a hill, or mass rotating on the string aka pendulum style etc but I am having very hard time explaining the solution to somebody else.

For example, if the centripetal force is towards the center, why do I feel like I am being thrown out of the circle? If I am twirling a mass at the end of the rope and if the rope were to break, does the mass really go in the tangential direction rather than the radially out?

I am able to explain some of this by saying that when the object is rotating with constant angular velocity, the free body diagrams has to show net force equal to the centripetal force i.e. net force is NOT zero. Now comes the crucial point to understand, somebody has to provide that force! It could be string tension or the track of the roller coaster. What you feel when on the roller coaster or turkish twist is given by the Newton's second law i.e. equal and opposite force. It is still difficult concept and if I can not explain it well means I do not quite get it.

Using the above explanation (aka tension in the string or force exerted by the roller coaster track) where is the physical force when airplane is doing vertical loop in the air?

 

[Shannow]

You don't have a physical force, you have a body in motion that wants to remain travelling in a straight line unless acted on by a force.

Take yourself travelling down a straight road, then taking a sharp left. Your head wants to keep travelling straight, but your body is dragged around by the car. The centripetal force provided by your neck drags your head around with it.

As to what happens when the rope breaks, I can guarantee that it scoots in exactly the direction it was travelling before the thread broke (i.e. the tangent to the curve at the point of breakage).

 

[Vikas]

I can draw the free body diagram and get the correct answer to the turkish twist problem but can you tell me which direction the normal force is acting so that generated frictional force is equal to the weight of the person stuck to the wall of the twister? Doesn't the normal force has to act to push against a surface to generate the frictional force? Why am I equating the normal force to the centripetal force? Which principle am I using there?

 

[Vikas]

Since we have some extremely talented people here, somebody can help on this one. The answer I got does not match my intuitive concepts. I got barely any temperature rise and I think there should have been some.

1.5 kg iron block is sliding on a rough surface at the velocity of 1.7 m/s. Assuming 71% of its kinetic energy ends up as frictional heat, how high the temperature of the block would rise when it comes to stop? Assuming specific heat of iron to be 0.450 J/gC

I got KE about 2.1675 J, 71% of that is 1.539 J Unfortunately, that does even make a dent in DeltaT :-( My answer? 0.002 C

Obviously, that is a wrong answer :-( I thought I was confusing J and g and kg but I don't think so. If I could get that energy number to be thousands times more, I would be able to get believable number. But 1 Newton already has a kg in it and 1 J needs 1 Newton, so I can not get the kilogram block moving at meter per second to be anything but single digit Joul in energy. But as far as heat is concerned, Joul needs grams to make any dent in the temperature and since our block has a mass of kilogram, it is like a massive heat sink.

 

[IndyIan]

Sounds right to me, 1.7 m/s is walking speed. And 1.5kg of iron isn't going to heat up at all, sliding to a stop from walking speed. Even sliding the block for an hour at 1.7m/s isn't going to heat it up alot once it gains some difference in temperature to ambiant.
Now if you want to calculate the surface temperature of the block, the instant it stops, that is something different, but still a very small temperature gain with such low energy.

 

[Vikas]

I am wondering if the said professor has screwed up some numbers here. But to be perfectly frank, I guess I too do not have real intuitive feel for Newton / Joul / Calorie / Kcal etc either.

 

[Vikas]

I just came across a very interesting but may be somewhat counter intuitive problem.

Suppose a a compact car is moving at 100 km/hr. So its kinetic energy is 2.8E5 J How many liters of room temperature water can be warmed from 20 C to boiling 100 C using this much energy?

I was completely unprepared for the answer!
































0.87 L

I thought it would have been lot more than that!

 

[Vikas]

Doing the iron block problem using dimensional analysis gives me the "right" answer.

Code:


KE = (1/2) mass (v)^2

Energy = sp_heat * mass * DeltaT



DeltaT = Energy / (sp_heat * mass )



(1/2)(mass)(v^2) * (71%)

DeltaT = -----------------

sp_heat * (mass)



mass cancels out, v=1.7 m/s, and sp_heat = 0.450 J/g.C



0.5 * ((1.7)^2) * 0.7

DeltaT = --------------------

0.450



DeltaT = 2.28 C

Comments? Even though this answer looks more "right", I think it is the wrong one.

 

[Shannow]

Nope, you've cancelled the mass of the car and the mass of the water...you've heated up a mass of water eqivalent to the car by 2.28C

the problem states that the energy is 280KJ (731Kg car).

280KJ/(4.2*80) gives .83KG of water, or 830ml.

Your specific heat at 4.5 is a bit high, that's the specific heat inside a boiler etc., not around ambient water temperatures.

My Y12 physics teacher had us do a calculation on the height that a block of Ice at zero C had to be dropped from to melt on impact...that's closer to your dimensionless number calcs.

 

[jrustles]

That is one fast frikkin Caddy!!

 

[Vikas]

RE: Shannow

830ml of water; that is the correct answer! For our friends who don't get the SI units, that is less than 1 quart for water!!

I do not know about you but I find it fascinating that a car barreling down the highway has the same energy as the quart of heated water! Think if you had to absorb the energy of that car aka the car colliding with a concrete wall; that would be the energy released in that incident. That energy is not even enough to boil a liter of water from room temperature!

As far as having a "bad day" is concerned, would you believe me that standing in front of the 60 mph car is not as bad as getting splashed by liter of boiling water?

How do you get around the fact that both these events release the same amount energy?

By the way, the original problem indeed had .002 degree increase in the iron block temperature as the answer. The more real sounding answer of 2 degree is indeed incorrect.

 

[Shannow]

Vikas, just to rattle some numbers, consider in a power station, over half a tonne per second of water is taken from 110F and 1psi (abs) to 2600psi, and 1000+F...there's millions of hp equivalents in that task...Like you say, the 830ml is interesting, but it then takes more energy to evaporate that same 830ml than it did getting it from 20C-100C.


I'd rather be hit by a flask of near boiling water than said car...m1v1+m2v2=m3v3 tells you how much of that energy is going to be transferred to you instantly