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Q60PE

Expert-verifiedFound in: Page 263

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Calculate the force the woman in Figure 7.46 exerts to do a push-up at constant speed, taking all data to be known to three digits. **

**(b) How much work does she do if her center of mass rises 0.240 m? **

**(c) What is her useful power output if she does 25 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.**

**Figure 7.46 Forces involved in doing push-ups. The woman’s weight acts as a force exerted downward on her center of gravity (CG).**

(a) The force the women exert to do a push-up is$490N$ .

(b) The work done to rise its center of mass is $117.6J$ .

(c) Her power output is $98W$ .

**The force exerted by the gravity on the body is known as the weight of the body. The weight of the body is measured in terms of newtons (N). **

Mathematically,

$w=mg$

Here, body's mass is m, and the acceleration due to gravity is g.

(a)

The net force acting on women is,

${\mathbf{F}}_{reaction}-\mathbf{w}=ma$

Here, ${F}_{reaction}$ is the force required by the women to perform push-up, w is the weight of the women, m is the mass of the women , and a is the acceleration of the women ($a=0$ as the women is performing push-up at constant speed).

Putting acceleration,

$\begin{array}{rcl}{\mathbf{F}}_{reaction}-\mathbf{w}& =& 0\\ {\mathbf{F}}_{reaction}& =& \mathbf{w}\\ & =& mg\end{array}$

Here, m is the mass of the women $\left(m=50.0kg\right)$ , and g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$ .

Putting all known values,

$\begin{array}{rcl}{\mathbf{F}}_{reaction}& =& \left(50.0\text{kg}\right)\times \left(9.8\text{m}/{\text{s}}^{2}\right)\\ & =& 490\text{N}\end{array}$

Therefore, the force the women exert to do a push-up is $490N$ .

(b)

The work done in rising center of mass is,

$W={\mathbf{F}}_{reaction}d$

Here,${\mathbf{F}}_{reaction}$ is the force exerted by women to perform push-up, and d is the height by which center of mass is raised $\left(d=0.240\text{m}\right)$ .

Putting all known values,

$\begin{array}{rcl}W& =& \left(490\text{N}\right)\times \left(0.240\text{m}\right)\\ & =& 117.6\text{J}\\ & & \end{array}$

Therefore, the work done to rise its center of mass is $117.6J$.

(c)

25 push-up means the women is doing 50 times work (25 times to push her up and 25 times to push her down). Therefore, the total work done in 25 push-up is,

${W}_{t}=50W$

Here, W is the work done to push her up or to push her down $\left(W=117.6\text{J}\right)$.

Putting all known values,

$\begin{array}{rcl}{W}_{t}& =& 50\times \left(117.6\text{J}\right)\\ & =& 5880\text{J}\\ & & \end{array}$

The power output of the women is,

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

Here, ${W}_{t}$ is the total work done to perform 25 push-up $\left(W=5880\text{J}\right)$, and t is the time taken to perform 25 push-up $\left(t=1\text{min}\right)$ .

Putting all known values,

$\begin{array}{rcl}P& =& \frac{5880\text{J}}{1\text{min}}\\ & =& \frac{5880\text{J}}{1\text{min}\times \left(\frac{60\text{s}}{1\text{min}}\right)}\\ & =& 98\text{W}\\ & & \end{array}$

Therefore, the power output is$98W$.

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