Chapter 10, Problem 96A

(a)

To determine

The average power produced for the first second.

Answer to Problem 96A

The average power produced for the first second is $1.5\times {10}^3\text{ W}$ .

Explanation of Solution

Given:

The mass of sprinter is $m=75\text{ kg}$ .

The total distance of the race is $d_{\text{f}}=50.0\text{ m}$ .

The time period is $t=8.5\text{s}$ .

Formula used:

The expression for the work done is,

$W=Fd$

Here, $F$ is the force applied and $d$ is the displacement.

The expression for the power is,

$P=\frac{W}{t}$

Here, $t$ is the time period.

Calculation:

Consider the initial displacement and the initial velocity as zero.

The final velocity is,

$\begin{array}{l}{v}_{\text{f}}=v_{\text{i}}+at\\ v_{\text{f}}=a{t}_1\end{array}$

The acceleration of the sprinter using the relation is,

$\begin{array}{l}{d}_{\text{f}}=\frac{1}{2}a{\left(t_1\right)}^2+v_{\text{f}}{t}_2\\ 50.0\text{ m}=\frac{1}{2}a{\left(1.0\text{ s}\right)}^2+a{t}_1{t}_2\\ 50.0\text{ m}=a\left(\frac{1}{2}{\left(1.0\text{ s}\right)}^2+\left(1.0\text{ s}\right)\times \left(\text{7}\text{.5 s}\right)\right)\\ a=6.25\text{m}/{\text{s}}^2\end{array}$

The displacement of the sprinter for the first second is,

$\begin{array}{l}d=\frac{1}{2}a{t}^2\\ d=\frac{1}{2}\left(6.25\text{m}/{\text{s}}^2\right)\times {\left(1.0\text{ s}\right)}^2\\ d=3.125\text{ m}\end{array}$

The average power produced for the first second is,

$\begin{array}{l}P=\frac{W}{t}\\ P=\frac{Fd}{t}\\ P=\frac{mad}{t}\\ P=\frac{\left(75\text{ kg}\right)\times \left(6.25\text{m}/{\text{s}}^2\right)\times \left(\text{3}\text{.125 m}\right)}{\left(\text{1}\text{.0 s}\right)}\\ P=1.5\times {10}^3\text{ W}\end{array}$

Conclusion:

Thus, the average power produced for that first second is $1.5\times {10}^3\text{ W}$ .

(b)

To determine

The maximum power generated by the sprinter.

Answer to Problem 96A

The maximum power generated by the sprinter is $3\times {10}^3\text{ W}$ .

Explanation of Solution

Given:

The mass of sprinter is $m=75\text{ kg}$ .

The displacement is $d=50.0\text{ m}$

The time period is $t=8.5\text{s}$ .

Calculation:

From part (a), the sprinter’s average power for the race is $P_{\text{avg}}=1.5\times {10}^3\text{ W}$ .

The maximum power developed by the sprinter is,

$\begin{array}{l}{P}_{\max }=2P_{\text{avg}}\\ P_{\max }=2\times \left(1.5\times {10}^3\text{ W}\right)\\ P_{\max }=3\times {10}^3\text{ W}\end{array}$

Conclusion:

Thus, the maximum power generated by the sprinter is $3\times {10}^3\text{ W}$ .