Q24P (a) Calculate the impulse experi... [FREE SOLUTION]

Chapter 7: Q24P (page 170)

(a) Calculate the impulse experienced when a 55-kg person lands on firm ground after jumping from a height of 2.8 m.
(b) Estimate the average force exerted on the person鈥檚 feet by the ground if the landing is stiff-legged, and again
(c) with bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm. [Hint: The average net force on him, which is related to impulse, is the vector sum of gravity and the force exerted by the ground. See Fig. 7鈥�34.] We will see in Chapter 9 that the force in (b) exceeds the ultimate strength of bone (Table 9鈥�2).
FIGURE 7-34 Problem 24.

Short Answer

Expert verified

(a) The magnitude of the impulse is \(407.55\;{\rm{N}} \cdot {\rm{s}}\).

(b) The average force exerted on the person鈥檚 feet by the ground if he lands with stiff legs is \(1.52 \times {10^5}\;{\rm{N}}\).

(c) The average force exerted on the person鈥檚 feet by the ground if he lands with bent legs is \(3.56 \times {10^5}\;{\rm{N}}\).

Step by step solution

Define potential energy

The potential energy is the energy associated with forces that depend on the position of the configuration of the particles.

It relies on the particle鈥檚 mass, gravitational acceleration, and the particle鈥檚 height from a reference point.

Given information

The mass of the person is \(m = 55\;{\rm{kg}}\).

The initial height of the person is \({h_1} = 2.8\;{\rm{m}}\).

The distance for stiff legs is \({d_1} = 1.0\;{\rm{cm}}\).

The distance for bent legs is \({d_2} = 50\;{\rm{cm}}\).

The final height of the person is \({h_2} = 0\) (taking ground as reference).

Calculate the magnitude of the impulse

(a)Calculate the velocity of the person by applying the conservation of energy principle.

\(\begin{array}{c}KE = - \Delta PE\\\frac{1}{2}m{v^2} = - mg\left( {{h_2} - {h_1}} \right)\\{v^2} = - 2g\left( {{h_2} - {h_1}} \right)\\v = \sqrt { - 2g\left( {{h_2} - {h_1}} \right)} \end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}v = \sqrt { - 2\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.\\} {{{\rm{s}}^{\rm{2}}}}}} \right)\left[ {\left( 0 \right) - \left( {2.8\;{\rm{m}}} \right)} \right]} \\v = 7.41\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

The magnitude of the impulse can be calculated as shown below:

\(\begin{array}{c}\left| J \right| = \left| {{p_f} - {p_i}} \right|\\\left| J \right| = \left| {0 - {p_i}} \right|\\J = mv\\J = \left( {55\;{\rm{kg}}} \right)\left( {7.41\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)\\J = 407.55\;{\rm{N}} \cdot {\rm{s}}\end{array}\)

Thus, the magnitude of the impulse is \(407.55\;{\rm{N}} \cdot {\rm{s}}\).

Calculate the average force exerted on the person’s feet by the ground if he lands with stiff legs

(b)

The average velocity can be calculated as shown below:

\(\begin{array}{l}{v_{avg}} = \frac{v}{2}\\{v_{avg}} = \frac{{\left( {7.41\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)}}{2}\\{v_{avg}} = 3.71\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

The time for collision with stiff legs can be calculated as shown below:

\(\begin{array}{c}\Delta t = \frac{{{d_1}}}{{{v_{avg}}}}\\\Delta t = \frac{{\left( {1\;{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)}}{{\left( {3.71\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)}}\\\Delta t = 2.69 \times {10^{ - 3}}\;{\rm{s}}\end{array}\)

The average force for stiff legs can be calculated as shown below:

\(\begin{array}{l}F = \frac{J}{{\Delta t}}\\F = \frac{{\left( {407.55\;{\rm{N}} \cdot {\rm{s}}} \right)}}{{\left( {2.69 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\F = 151.51 \times {10^3}\;{\rm{N}}\end{array}\)

The net average force is

\(\begin{array}{l}{F_{net}} = F + mg\\{F_{net}} = \left( {151.51 \times {{10}^3}\;{\rm{N}}} \right) + \left( {55\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}}\right.\\}{{{\rm{s}}^{\rm{2}}}}}}\right)\\{F_{net}}=1.52\times {10^5}\;{\rm{N}}{\rm{.}}\end{array}\)

Thus, the average force exerted on the person鈥檚 feet by the ground if he lands with stiff legs is \(1.52 \times {10^5}\;{\rm{N}}\).

Calculate the average force exerted on the person’s feet by the ground if the landing is with bent legs

(c)The time for collision for bent legs can be calculated as shown below:

\(\begin{array}{l}\Delta t=\frac{{{d_2}}}{{{v_{avg}}}}\\\Delta t=\frac{{\left({50\;{\rm{cm}}}\right)\left({\frac{{{{10}^{2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}\right)}}{{\left({3.71\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}} \right.\\} {\rm{s}}}} \right)}}\\\Delta t = 134.77 \times {10^{ - 3}}\;{\rm{s}}\end{array}\)

The average force for bent legs can be calculated as shown below:

\(\begin{array}{l}F = \frac{J}{{\Delta t}}\\F = \frac{{\left( {407.55\;{\rm{N}} \cdot {\rm{s}}} \right)}}{{\left( {134.77 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\F = 3.02 \times {10^3}\;{\rm{N}}\end{array}\)

The net average force is

\(\begin{array}{l}{F_{net}}= F + mg\\{F_{net}} = \left( {3.02 \times {{10}^3}\;{\rm{N}}} \right) + \left( {55\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}}\mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. \\}{{{\rm{s}}^{\rm{2}}}}}} \right)\\{F_{net}} = 3.56 \times {10^5}\;{\rm{N}}{\rm{.}}\end{array}\)

Thus, the average force exerted on the person鈥檚 feet by the ground if he lands with bent legs is \(3.56 \times {10^5}\;{\rm{N}}\).