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3 years ago

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A 100-kg athlete is performing a vertical leap to test his leg strength. He leaves the ground at 3.20 m/s, zooms upward to some

maximum altitude, and then returns to the ground. As he lands, he bends his legs and crouches down, so that he moves 30 cm downward as he comes to a stop. What average force did the athlete’s legs exert on the ground as he landed?

1 answer:

MaRussiya [10]3 years ago

5 0

Answer:

2687.7 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

When he will reach the ground his speed will be 3.2 m/s but will travel 30 cm down

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.2^2}{2\times 0.3}\\\Rightarrow a=-17.067\ m/s^2$

Force

$F=m(g-a)\\\Rightarrow F=100\times (9.81-(-17.067))\\\Rightarrow F=2687.7\ N$

The force on the legs will be 2687.7 N

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Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

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$F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (Electric force of q₁ over q₃)

Known data

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Problem development

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$\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (We cancel k and q₃)

$\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}$

q₂(2-x)² = q₁x²

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6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

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9x² + 24x - 24 = 0

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