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Question

# Biomechanical research has shown that when a 67-kg person is running, the force exerted on each foot as it strikes the ground can be as great as $2300 \mathrm{~N}$. If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person's weight, what are the magnitude and direction of the person's acceleration?

Solution

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$\textbf{Given values:}$

$m=67 \: \text{kg}$

$F=2300 \: \text{N}$

$g=9.81 \: \text{m}/\text{s}^2$

So, the net force acting on the person is given by :

$\sum F_{y}=m \cdot a_{y}$

From previous equation, we have to find value for $a_{y}$ :

\begin{align*} F+W&=m \cdot a_{y}\\ F+(m \cdot g)&=m \cdot a_{y}\\ 2300 \: \text{N}+(67 \: \text{kg}) \cdot (9.81 \: \text{m}/\text{s}^2)&=(67 \: \text{kg}) \cdot a_{y} \tag{Substitute values in eq.}\\ 2300 \: \text{N}+(-657.27 \: \text{N})&=(67 \: \text{kg}) \cdot a_{y}\\ 2300 \: \text{N}-657.27 \: \text{N}&=(67 \: \text{kg}) \cdot a_{y}\\ 1642.73 \: \text{N}&=(67 \: \text{kg}) \cdot a_{y}\\ a_{y}&=\frac{1642.73 \: \text{N} }{67 \: \text{kg}}\\ a_{y}&=\frac{1642.73 \: \text{kg} \cdot \text{m}/\text{s}^2 }{67 \: \text{kg}}\tag{1 \: \text{N}=1 \: \text{kg} \cdot \text{m}/\text{s}^2.}\\ a_{y}&=24.518 \: \text{m}/\text{s}^2\\ \end{align*}

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