A 100 kg person stands on a scale. a.) What would be the scale readout? N = · mg
ID: 1424646 • Letter: A
Question
A 100 kg person stands on a scale.
a.) What would be the scale readout? N = · mg (Express in Newtons, and as a multiple of the person's weight.)
Now, find a general expression for the scale readout, in terms of the person's mass, acceleration, and/or g. Show that it can be used to get the answer for part a above, and then use it to get the answers for all of the following parts:
b.) If the person stands on the scale in an elevator accelerating upwards at 2 m/s2, what is the scale readout? N = · mg (Does the person feel heavier or lighter than the "usual" of part a?)
c.) If the person stands on the scale in an elevator accelerating downwards at 2 m/s2, what is the scale readout? N = · mg (Does the person feel heavier or lighter than usual?)
d.) If the person stands on the scale in an elevator accelerating downwards at 9.8 m/s2, what is the scale readout? N = · mg (Does the person feel heavier or lighter than usual?)
e.) In parts b-d, was the actual weight of the person different? Explain the difference between apparent weight and actual weight.
f.) If the person stands on the scale in an elevator moving up at a constant speed of 100 m/s, what is the scale readout? N = · mg (Does the person feel heavier or lighter than usual?) Explain why during most of the time during an elevator ride, a person will perceive nothing unusual.
Explanation / Answer
a)
scale reading N = mg = 100*9.8 = 980 N
b)
Fnet = N - mg
but from newtons second Fnet = ma
N - mg = m*a
N = mg + ma = 980 + (100*2) = 1180 N
(c)
Fnet = N mg
but from newtons second Fnet = -ma
N - mg = -m*a
N = mg - ma = 980 - (100*2) = 780 N
(d)
N = mg - ma = mg - mg = 0
(e)
In b apparent is more than actual weight
In c apparent is less than actual weight
In d apparent is same as actual weight
(f)
for constant speed acceleration a = 0
N = mg = 980 N
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