Physics problem to be solved please

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Physics problem to be solved please

Postby Bill321 » Fri Nov 18, 2005 9:46 am

A 60-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when:
a) the elevator is descending at a constant rate of 10 m/s
b) the elevator is accelerating downward a 2 m/s^2
c)the elevator is ascending and its speed is decreasing at the rate of 2 m/s^2, and
d) the elvator is accelerating upward at 2 m/s^2
Bill321
 

Postby shill » Fri Dec 23, 2005 6:16 am

If you want to do these problems on your own, read this page.




The acceleration due to gravity is 9.80665 m/s² [down].

a) Since the elevator is descending at a constant rate, the acceleration is zero. The net force will be 588.399 N, and the scale will read 60 kg.

b) (m*a)+f_gravity = (60. kg * -2.0 m/s²) + 588.399 N = 468.3993 N. Divide by acceleration due to gravity, therefore the scale will read approximately 47.8 kg.

c) This one is a little trickier. The elevator is ascending, but it is slowing down, so the acceleration is going in the opposite direction (downward). Since this acceleration (2.0 m/s² [down]) is exactly the same as in the previous question, the scale will read the same (47.8 kg).

d) Acceleration is 2.0 m/s² [up]. Use the same formula as in question b.
(m*a)+f_gravity/g = ((60. kg * 2.0 m/s²) + 588.399 N)/9.80665 m/s²) = approximately 72.2 kg.

I make no claims to the correctness of this advice. Use your own judgement to confirm this advice.
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Postby Guest » Fri Dec 23, 2005 7:19 am

Mostly to be a smartass, note that we haven't established WHAT kind of a scale it is.

If it is a "dishonest weight, chock full of springs" scale, I agree with shill's method (I didn't check all the figures).

If it is a physicians' scale, or any type of "beam balance" scale, the counterweights will be equally affected as the girl, and the scale will indicate 60 kg. (the sensitivity may be slightly affected as "gravity" is higher or lower by about 20%)
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Postby Potamus » Tue Jan 10, 2006 11:02 am

Do your own homework. :roll:
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