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Sindrei [870]
3 years ago
6

An athlete of mass 70.0 kg applies a force of 500 N to a 30.0 kg luge, which is initially at rest, over a period of 5.00 s befor

e jumping onto the luge. Assuming there is no friction between the luge and the track on which it runs, what is its velocity after the athlete jumps on?
Physics
1 answer:
ira [324]3 years ago
6 0
Force = (mass) x (acceleration)

Acceleration = (force) / (mass)

Acceleration of the luge = (500 N) / (30 kg) = 16 and 2/3 m/s²

After accelerating at that rate for 5 sec, its speed is (16-2/3) x (5) = 83-1/3 m/s .

(I pause slightly at this point, to reflect that this thing is now moving
at about 186 miles per hour.  I question that, and I check my work.
I reassure myself with two thoughts:  1). Maybe those things really
do move at that kind of speeds.  I don't know.  2). I was given the
numbers, and I didn't make them up, so I'm only responsible for the
math, not for the plausibility of the solution.) 

So the luge is moving at 83-1/3 m/s when he jumps on.  In order to maintain
that force against it for 5 seconds, he had to accelerate himself to almost-
if-not-totally the same speed ... necessary, no matter how implausible.

So, although it hasn't been mentioned, the pusher is also doing an enormous
amount of other work just to accelerate himself, and when he jumps aboard,
his own velocity already matches that of his luge.  I'm going to say that
after the jump, they continue on, together, coupled as one, at the same
speed as just before the jump.

Their speed together is <em>83-1/3 m/s</em> .

We can't state their <em>velocity</em>, because no information is given regarding
the direction of the track.
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k is the constant of proportionality

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substituting (1) into (2)

                                     I = \frac{k}{s} \int\limits\, dS

                                     I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS

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                                     I = 2\pi ka

                                     k = \frac{I}{2\pi a}

substitute J = \frac{k}{s}

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