about 1/10th as great by bending your legs
Time extended decreases acceleration value.
solve using Newton's 2nd and Uniform Acceleration Laws. plug in values for descending.
Vf = 0.0 m/s
Vi = 10 m/s
t = 1 s
Acceleration is given by
a = [Vf - Vi] / t
a = [ (0.0 m/s) - (10 m/s) ] / (1 s)
a = [ -10 m/s ] / (1 s)
a = -10 m/s^2
Solve the same thing, but with time 10x, so t = 10 s
a = [ (0.0 m/s) - (10 m/s) ] / (10 s)
a = [ -10 m/s ] / (10 s)
a = -1 m/s^2
Solve for first force and second force, using a mass of 100 kg
F = m * a
F = (100 kg) * (-10 m/s^2)
F = -1,000 N
F = (100 kg) * (-1 m/s^2)
F = -100 N
So divide long-force by short-force
(-100 N) / (-1,000 N) = 0.1 (which is 1/10)
The average force of person with bending legs becomes 1/10 times the force exerted with stiff legs.
Let m be the mass of a person.
Now, the expression for the average force offered by the person at standing condition is,
[tex]F_{1}=\dfrac{mv}{t_{1}}[/tex] .........................................................(1)
Here, v is the speed of person and [tex]t_{1}[/tex] is the impact time.
Now, consider the bending condition,, where the impact time is about 10 times as great as for a stiff legged landing. Which means,
[tex]t_{2}=10 \times t_{1}[/tex]
Then, average force is,
[tex]F_{2}=\dfrac{mv}{t_{2}} \\\\F_{2}=\dfrac{mv}{10 \times t_{1}} ............................................................(2)[/tex]
Taking ratio of equation (1) and (2) as,
[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{\dfrac{mv}{t_{1}}}{\dfrac{mv}{ 10 \times t_{1}}} \\\\F_{1} = 10 \times F_{2}\\\\F_{2}=\dfrac{F_{1}}{10}[/tex]
Thus, we can conclude that on bending the legs, the average force of person becomes 1/10 times the force exerted with stiff legs.
Learn more about the average force here:
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