1.i. If the KE of a particle is doubled, by what factor has its speed
increased?


ii.If the speed of a particle is doubled, by what factor does its KE
increase?
 
2.i. How much work must be done to stop a 1000 kg car travelling at
100 km/h?


ii.If the car stops in a distance of 193 m, determine the average
force which stopped the car.

Here again pj4444!!!

As my fellow researcher safe-ga did, I will supose that KE means
Kinetic energy.

safe-ga did a good job with the part 1:

KE = 1/2 * m * V^2

then given the KE, we have:

V1 = sqrt(2 * KE / m)          ,sqrt means square root.

If the KE is doubled, then:

V = sqrt(2 * 2 * KE / m) = sqrt(2) * sqrt(2 * KE / m) = sqrt(2) * V1 ;

The initial speed V1 is increased by a factor sqrt(2) = 1.4142


Now we have the speed given V, then:

KE1 = 1/2 * m * V^2

If the speed is doubled:

KE = 1/2 * m * (2V)^2 = 1/2 * m * 2^2 * V^2 = 4 * (1/2 * m * V^2) = 4
* KE1

The initial KE is increased by a factor of 4.

This finished the part 1.

-------------------------------------------------

2.i

First of all the units

100km/h = 100,000m/3600s = 27.78m/s

The initial KE is:

KE1 = 1/2 * 1000 * (27.78^2) = 385,864.2 J

The final KE is zero.

Then we "loss" all the Kinetic energy of the car, what happened
here!!!???

Some non-conservative forces did a work which results in a stopped
car.

This "non-conservative work" (Wnc) is calculated in this case:

Wnc = KE1 - KEf = 385,864.2 - 0 = 385,864.2 J.

This anser 2.i

Note: going further, the Mechanic energy (E) is used to calculate Wnc;
E = KE + PE + EE (Kinetic + Potential + Elastic), and the Wnc is:
Wnc = Ef - E0 . 
In this problem we have neither Potential nor Elastic Energies, so the
problem is reduced to calculate the KE variation.
End of note.


2.ii

W = F * d ; where d is the distance over an average force is applied.
If the car stops in a distance of 193 m, the average force which
stopped the car is:
F = Wnc / d = 385,864.2 / 193 = 1,999.3 N .(this force could be
brakes, friction or both)



Again I hope this helps you, if you need a clarification, please post
a request for it.

Thank you.
livioflores-ga

pj,

I'm not a native English speaker, so I might get this wrong and my
notations might differ from those you are used to. However, I'll try
to answer your questions.

1.i. I suppose KE means Kinetic Energy, which is equal to the product
of the mass and the square of the spped divided by 2. That is:

KE = m*s*s/2  (if m stands for mass and s for speed)

As a result, if you double the KE (and suppose the mass stays the
same) you'll have something like this:

2*KE = m*s'*s'/2  (where s' is the new speed)

From the two relations above you get

s'*s' = 2*s*s

So, supposing that both s and s' are positive(which they are) you get
that s' equals s multiplied by the square root of 2. The answer is:
the speed is multiplied by the square root of 2.


1.ii. By doing a similar calculation to the one above, the result is 4
(i.e. if you double the speed, the KE grows by a factor of 4).

Hope this helps,
SAfe.

The answer to 2.i. is almost right, but not quite.  The amount of work
required to stop a moving car is negative, because the force is
applied in the opposite direction of the motion.  So the answer has
the same magnitude as that given by livioflores, but is negative.