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Expert-verified Found in: Page 263 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.) (b) Compare this force with the weight of the jogger.

(a) The magnitude of the force needed to stop the downward motion of a jogger’s leg is $$16335{\rm{ N}}$$.

(b) The ratio of the total force to the weight of the jogger is $$22.2$$ times.

See the step by step solution

## Step 1: Weight

Weight is a force which acts on a mass due to acceleration due to gravity.

Mathematically,

$$W = mg$$

Here, m is the mass of the object, and g is the acceleration due to gravity.

## Step 2: Magnitude of the force

(a)

The force on leg is,

\begin{aligned}{c}F = {m_l}a\\ = \frac{{{m_l}{v^2}}}{{2s}}\end{aligned}

Here, $${m_l}$$ is the mass of the leg $$\left( {{m_l} = 13.0{\rm{ kg}}} \right)$$, v is the velocity $$\left( {v = 6.00{\rm{ m}}/{\rm{s}}} \right)$$ and s is the stopping distant $$\left( {s = 1.50{\rm{ cm}}} \right)$$.

Putting all known values,

\begin{aligned} F &= \frac{{\left( {13.0{\rm{ kg}}} \right) \times {{\left( {6.00{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( {1.50{\rm{ cm}}} \right)}}\\ &= \frac{{\left( {13.0{\rm{ kg}}} \right) \times {{\left( {6.00{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( {1.50{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}\\ &= 15600{\rm{ N}}\end{aligned}

The weight of the body is,

$${W_g} = {m_b}g$$

Here, $${m_b}$$ is the mass of the body $$\left( {{m_b} = 75{\rm{ kg}}} \right)$$, and g is the acceleration due to gravity $$\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)$$.

Putting all known values,

\begin{aligned} {W_g} &= \left( {75.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ &= 735{\rm{ N}}\end{aligned}

The total force on the body is,

$${F_{total}} = F + {W_g}$$

Putting all known values,

\begin{aligned} {F_{total}} &= \left( {15600{\rm{ N}}} \right) + \left( {735{\rm{ N}}} \right)\\ &= 16335{\rm{ N}}\end{aligned}

Therefore, the magnitude of the force needed to stop the downward motion of a jogger’s leg is $$16335{\rm{ N}}$$.

## Step 3: Compare the ratio of the force to the weight

(b)

The ratio of the total force to the weight is,

\begin{aligned} \frac{{{F_{total}}}}{{{W_g}}} &= \frac{{16335{\rm{ N}}}}{{735{\rm{ N}}}}\\ &= 22.2\end{aligned}

Therefore, the ratio of the total force to the weight is $$22.2$$. ### Want to see more solutions like these? 