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Q53PE

Expert-verifiedFound in: Page 263

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.

**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.

(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 \(16335{\rm{ N}}\).

(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 \(22.2\).

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