# Net Force Question

#### student34

1. Homework Statement

75.0kg Man decends from 3.1m above the ground. At 3.1m his feet touch the ground and he absorbs the landing through a constant acceleration of 0.6m to come to a verticle rest. What is the net force on him?

2. Homework Equations

Vi = √(2*a*y); a = -[(Vi)^2]/(-2*y); F = m*a; where y = -0.60m and a = -9.8m/s^2.

3. The Attempt at a Solution

Vi = √[2*a*(-0.60m)] = -7.79m/s
a = -[(-7.79m/s)^2]/(-2*(-0.60m) = -50.63m/s^2
F = 75.0kg*(-50.63m/s^2) = -3800N.

I get 3800N from the ground to him, but the book has 4530N. Their answer works if we add on the absolute value of his mass multiplied by gravity, but why would they do that?

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#### sankalpmittal

1. Homework Statement

75.0kg Man decends from 3.1m above the ground. At 3.1m his feet touch the ground and he absorbs the landing through a constant acceleration of 0.6m to come to a verticle rest. What is the net force on him?

2. Homework Equations

Vi = √(2*a*y); a = -[(Vi)^2]/(-2*y); F = m*a; where y = -0.60m and a = -9.8m/s^2.

3. The Attempt at a Solution

Vi = √[2*a*(-0.60m)] = -7.79m/s
a = -[(-7.79m/s)^2]/(-2*(-0.60m) = -50.63m/s^2
F = 75.0kg*(-50.63m/s^2) = -3800N.

I get 3800N from the ground to him, but the book has 4530N. Their answer works if we add on the absolute value of his mass multiplied by gravity, but why would they do that?

Firstly ground has to apply normal reaction R equal to his weight to balance him. Secondly, in this problem the person possesses additional force ma due to the kinematic equation. To make him come to vertical rest, ground must apply net force equal to what he possessed. In this case he had net force mg+ma. By newton's third law, normal reaction should be equal and opposite to it.

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