91Ó°ÊÓ

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 \(22.2\).