Chapter 7, Problem 68P

(a)

To determine

The internal work-done by the athlete.

Answer to Problem 68P

400 Joules

Explanation of Solution

Given:

The center of mass of athlete moves 0.500 meters.

The muscular energy that needs to be spent on the athlete to lift his body is 800 N.

Formula Used:

The work-done by the internal energy is

  $W=Force\times displacement$

Calculation:

800 N is basically the force that is needed to overcome by the gravity to jump upward.

  $W=800\times 0.500=400\text{ Joules}$

Conclusion:

The work-done by the internal energy of jumper is 400 Joules.

(b)

To determine

The speed of the athlete.

Answer to Problem 68P

  $v=3.16\text{m/s}$

Explanation of Solution

Given:

The center of mass of athlete moves 0.500 meters.

The muscular energy that needs to be spent on the athlete to lift his body is 800 N.

Formula Used:

The potential energy is equal to the kinetic energy of the athlete

  $\begin{array}{l}P.E=K.E\\ mgh=\frac{1}{2}m{v}^2\\ v=\sqrt{2gh}\end{array}$

Calculation:

Considering $g=10{\text{m/s}}^{\text{2}}$

  $\begin{array}{l}v=\sqrt{2gh}\\ =\sqrt{2\times 10\times 0.500}\\ =\sqrt{10}\text{ m/s}\end{array}$

  $v=3.16\text{m/s}$

Conclusion:

The speed is $v=3.16\text{m/s}$ .

(c)

To determine

The time of jump for given height.

Answer to Problem 68P

The time required to reach 0.400 m height by the jumper will be 0.5145 sec.

Explanation of Solution

Given:

Initial speed, $u=3.16\text{ m/s}$

Final speed, $v=0$

Formula Used:

  $v^2-u^2=2as$

  $\Rightarrow a=0-\frac{u^2}{2s}$

  $s=ut-\frac{1}{2}a{t}^2$

Calculation:

  $v^2-u^2=2as$

  $\Rightarrow a=0-\frac{u^2}{2s}$

  $a=-\frac{{3.16}^2}{2\times 0.500}=-\frac{10}{1}=-10{\text{m/s}}^{\text{2}}$

Negative sign shows the deceleration.

Now, the acceleration of the jumper is $-10{\text{m/s}}^{\text{2}}$ . The time required for center of mass of the jumper to reach height 0.400 m can be found using equation

  $s=ut-\frac{1}{2}a{t}^2$

  $0.400=3.16\times t-0.5\times 10\times t^2$

Solving the equation, t will be -0.1174 seconds.or 0.5145 seconds. Negative time does not make any sense. Hence, the t will be 0.5145 seconds.

Conclusion:

The time required to reach 0.400 m height by the jumper will be 0.5145 s.

(d)

To determine

The average mechanical power.

Answer to Problem 68P

  $1265.822\text{ Joule/s}$

Explanation of Solution

Given:

Initial speed, $u=3.16\text{ m/s}$ .

Final speed, $v=0$ .

Formula Used:

  $v^2-u^2=2as$

Calculation:

  $a=-\frac{u^2}{2s}$

  $\begin{array}{l}a=-\frac{{3.16}^2}{2\times 0.500}\\ =-\frac{10}{1}\\ =-10{\text{m/s}}^{\text{2}}\end{array}$

Negative sign shows the deceleration.

Now, the acceleration of the jumper is $-10{\text{m/s}}^{\text{2}}$ . The time required for center of mass of the jumper to reach height 0.500 m can be found using equation $v-u=at$

As, $v=0$ ,

  $t=-\frac{u}{a}=-\frac{3.16}{-10}=0.316\mathrm{s}$

  $Power=\frac{W.D}{t}=\frac{400}{0.316}=1265.822\text{ Joule/s}$

Conclusion:

The average power is 1265.822 Watts.