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PIT_PIT [208]
3 years ago
6

A 57 kg woman is on an elevator accelerating upward at "1.25" meters per second squared. what is the normal force acting on her?

Physics
1 answer:
ella [17]3 years ago
8 0

Answer:

Normal force acting on the woman is 629.85 N.

Explanation:

Given:

Mass of the woman is, m=57\textrm{ kg}

Net upward acceleration is, a=1.25\textrm{ }m/s^{2}

Acceleration due to gravity, g=9.8\textrm{ }m/s^{2}

Let the normal force acting upward be R newtons.

Therefore, net force in the upward direction is given as:

Net force = Upward force - Downward force.

Downward force acting on the woman is her weight which is equal to mg.

Therefore, Net force = R-mg

Now, as per Newton's second law of motion,

Net force, F_{net}=ma

So,

R-mg=ma\\R=mg+ma\\R=m(g+a)

Plug in 57 kg for m, 9.8 m/s² for g, and 1.25 m/s² for a. Solve for R. This gives,

R=57(9.8+1.25)\\R=57(11.05)=629.85\textrm{ N}

Therefore, the normal force acting on her is 629.85 N.

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