Chapter 5, Problem 22P

A 60.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.0 m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete’s speed is 9.0 m/s as y = 0. What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (e) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

To determine

The interacting objects and how do they interact.

Answer to Problem 22P

The interacting objects are the athlete and the earth and the interaction is gravitational force.

Explanation of Solution

When the athlete jumps and when he is in the air, the interacting objects are the athlete and the earth. The gravitational force between the athlete and the earth will be the interaction between the objects.

Thus, the interacting objects are the athlete and the earth and the interaction is gravitational force.

Conclusion:

The interacting objects are the athlete and the earth and the interaction is gravitational force.

To determine

The kinetic energy and the gravitational potential energy of the athlete at $h=0$ level.

Answer to Problem 22P

The kinetic energy of the athlete at $h=0$ level is $2.4\times {10}^3\text{J}$ and the potential energy of the athlete at $h=0$ level is $\text{zero}$.

Explanation of Solution

Section 1

To determine: The kinetic energy of the athlete at $h=0$ level.

Answer: the kinetic energy of the athlete at $h=0$ level is $2.4\times {10}^3\text{J}$.

Explanation:

Given Info:

The speed of the athlete at $h=0$ level is $9.0\text{m/s}$.

The mass of the object is $60.0\text{kg}$.

Formula to calculate the kinetic energy of the athlete is,

$K{E}_i=\frac{1}{2}m{v}_i{}^2$

• $v_i$ is the speed of the athlete at $h=0$ level
• m is the mass of the athlete.

Substitute $60.0\text{kg}$ for m and $9.0\text{m/s}$ for $v_i$ to find the kinetic energy of the athlete,

$\begin{array}{c}K{E}_i=\frac{1}{2}\left(60.0\text{kg}\right){\left(9.0\text{m/s}\right)}^2\\ =2.4\times {10}^3\text{J}\end{array}$

Thus, the kinetic energy of the athlete at $h=0$ level is $2.4\times {10}^3\text{J}$.

Section 2

To determine: The potential energy of the athlete at $h=0$ level.

Answer: The potential energy of the athlete at $h=0$ level is $\text{zero}$.

Explanation:

Given Info:

The speed of the athlete at $h=0$ level is $9.0\text{m/s}$.

The mass of the object is $60.0\text{kg}$.

Formula to calculate the potential energy of the athlete is,

${\left(P{E}_g\right)}_i=mg{h}_i$

• $h_i$ is the height of the level
• m is the mass of the athlete.

Substitute $60.0\text{kg}$ for m and $zero$ for $h_i$ to find the potential energy of the athlete,

$\begin{array}{c}{\left(P{E}_g\right)}_i=\left(60.0\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\left(0\right)\\ =0\end{array}$

Thus, the potential energy of the athlete at $h=0$ level is $\text{zero}$.

Conclusion:

The kinetic energy of the athlete at $h=0$ level is $2.4\times {10}^3\text{J}$ and the potential energy of the athlete at $h=0$ level is $zero$.

To determine

The kinetic energy and the gravitational potential energy of the athlete at maximum height.

Answer to Problem 22P

The kinetic energy of the athlete at maximum height is zero and the potential energy of the athlete at maximum height is $2.4\times {10}^3\text{J}$

Explanation of Solution

Section 1

To determine: The kinetic energy of the athlete at maximum height.

Answer: the kinetic energy of the athlete in the maximum height is zero.

Explanation:

Given Info:

The mass of the object is $60.0\text{kg}$.

Formula to calculate the kinetic energy of the athlete is,

$K{E}_f=\frac{1}{2}m{v}_f{}^2$

• $v_f$ is the speed of the athlete at the maximum height
• m is the mass of the athlete.

When the athlete reaches its maximum height in the jump, the athlete will be in the rest at the maximum height. Thus, the kinetic energy of the athlete will be zero in the maximum height.

Thus, the kinetic energy of the athlete in the maximum height is zero.

Section 2

To determine: The potential energy of the athlete at maximum height.

Answer: the potential energy of the athlete at maximum height is $2.4\times {10}^3\text{J}$.

Explanation:

Given Info:

The mass of the object is $60.0\text{kg}$.

When the athlete reaches the maximum height, according to conservation of energy; the total energy will be converted in the form of gravitational potential energy.

$\begin{array}{l}\Delta P{E}_g=-\Delta KE\\ \end{array}$

Thus,

$P{E}_f=P{E}_i+K{E}_i-K{E}_f$

• $P{E}_i$ is the initial potential energy
• $K{E}_i$ is the initial kinetic energy
• $K{E}_f$ is the final kinetic energy

Substitute $\text{zero}$ for $P{E}_i$, $2.4\times {10}^3\text{J}$ for $K{E}_i$, and $\text{zero}$ for $K{E}_f$ to find the potential energy of the athlete,

$\begin{array}{c}P{E}_f=0+2.4\times {10}^3\text{J}-0\\ =2.4\times {10}^3\text{J}\end{array}$

Thus, the potential energy of the athlete at maximum height is $2.4\times {10}^3\text{J}$.

Conclusion:

The kinetic energy of the athlete at maximum height is zero and the potential energy of the athlete at maximum height is $2.4\times {10}^3\text{J}$.

To determine

The general equation for energy conservation and the maximum height and find the numerical answer for maximum height.

Answer to Problem 22P

The maximum height of the jump of the athlete is $4.1\text{m}$.

Explanation of Solution

Since the athlete’s total energy is conserved,

$\Delta KE+\Delta PE=0$

Thus,

$K{E}_f+P{E}_f=K{E}_i+P{E}_i$

Hence, the general equation for energy conservation is;

$\frac{1}{2}m{v}_f{}^2+\frac{1}{2}mg{h}_f=\frac{1}{2}m{v}_i{}^2+mg{h}_i$

Formula to calculate the final height of the athlete is,

$\begin{array}{c}{h}_f=\frac{mg{h}_i+\frac{1}{2}m{v}_i{}^2-\frac{1}{2}m{v}_f{}^2}{mg}\\ =h_i+\frac{\left(v_i{}^2-v_f{}^2\right)}{2g}\end{array}$

Substitute $\text{zero}$ for $h_i$, $9.0\text{m/s}$ for $v_i$ , $\text{zero}$ for $v_f$ and $9.8{\text{m/s}}^{\text{2}}$ for g to find the final height,

$\begin{array}{c}{h}_f=0+\frac{{\left(9.0\text{m/s}\right)}^2-0}{2\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ =4.1\text{m}\end{array}$

Thus, the maximum height of the jump of the athlete is $4.1\text{m}$.

Conclusion:

The maximum height of the jump of the athlete is $4.1\text{m}$.

To determine

The general equation for energy conservation and the velocity at half the maximum height and find the numerical answer for the same.

Answer to Problem 22P

The velocity at half the maximum height of the athlete is $6.4\text{m/s}$.

Explanation of Solution

Since the athlete’s total energy is conserved,

$\Delta KE+\Delta PE=0$

Thus,

$K{E}_f+P{E}_f=K{E}_i+P{E}_i$

Hence, the general equation for energy conservation is;

$\frac{1}{2}m{v}_f{}^2+\frac{1}{2}mg{h}_f=\frac{1}{2}m{v}_i{}^2+mg{h}_i$

Formula to calculate the velocity at half the maximum height is,

$\begin{array}{l}{v}_f{}^2=\frac{2}{m}\left(\frac{1}{2}m{v}_i{}^2+mg{h}_i+-mg{h}_f\right)\\ v_f=\sqrt{v_i{}^2+2g\left(h_i-h_f\right)}\end{array}$

Substitute $zero$ for $h_i$, $9.0\text{m/s}$ for $v_i$ , $h_{\max }/2$ for $h_f$ and $9.8{\text{m/s}}^{\text{2}}$ for g to find the final speed,

$\begin{array}{c}{v}_f=\sqrt{{\left(9.0\text{m/s}\right)}^2+2\left(9.8{\text{m/s}}^2\right)\left(0-\frac{4.1\text{m}}{2}\right)}\\ =6.4\text{m/s}\end{array}$

Thus, the velocity at half the maximum height of the athlete is $6.4\text{m/s}$.

Conclusion:

The velocity at half the maximum height of the athlete is $6.4\text{m/s}$