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If the mass of the paper is 0.003 kg, what force does the boxer except on it?​

If the mass of the paper is 0.003 kg, what force does the boxer except on it?​-example-1

1 Answer

3 votes

Answer:

1.15 N

Step-by-step explanation:

You want to know the force exerted on a mass of 0.003 kg to accelerate it from 0 to 23 m/s in a period of 0.06 s.

Acceleration

The acceleration of the mass is the change in velocity divided by the change in time:

a = ∆v/∆t

a = ((23 -0) m/s)/(0.06 s) = 1150/3 m/s²

Force

The force applied is the product of mass and acceleration:

F = ma

F = (0.003 kg)(1150/3 m/s²) = 1.15 kg·m/s² = 1.15 N

The applied force is 1.15 newtons.

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