Suppose your mass is 75.0 kg. You step gently off the edge of a high diving plat

Question

Suppose your mass is 75.0 kg. You step gently off the edge of a high diving platform, so that you essentially start falling from rest toward the Earth. (Ignore air resistance.) a. During your freefall, which force has the greater magnitude? A. the magnitude of the gravitational force of the Earth on you B. the magnitude of the gravitational force of you on the Earth C. neither - they are equal b. During freefall, your downward acceleration will have a magnitude of 9.80 m/s^2. At the same time, what is the magnitude of the Earth's upward acceleration ? Show your work.

Explanation / Answer

The weight of the teacher = mg = 75x10 = 750N

Remember F=ma. If velocity is constant that means a=0. So the resultant force F = m x 0 = 0.

Call W the downwards gravitational pull of the earth on the person (the person's weight).
Call N the upwards normal contact force (reaction) of the scales on the person (i.e the 'support' provided by the scales).
The resultant force on the person (using up as positive) is F = N - W


When velocity is constant, F = 0 so 0 = N - W. That means N = W. So N =750 N as you would expect.

Your diagram should have:
Downwards arrow marked: Weight (gravitational pull of earth on person) = 750N down
Upwards arrow marked: Normal force of scales on person) = 750N up
The arrows are equal lengths

so option A

b.

In a uniform gravitational field, in the absence of any other forces, gravitation acts on each part of the body equally and this isweightlessness, a condition that also occurs when the gravitational field is zero (such as when far away from any gravitating body). A body in free fall experiences "0 g"

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