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### College Physics (Urone)

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 $98W$ .

See the step by step solution

## Step 1: Weight

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.

## Step 2: Net Force acting on the woman

(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,

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

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

## Step 3: Work done to rise its center of mass

(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,

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

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

## Step 4: Useful power output

(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,

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

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,

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

Therefore, the power output is$98W$.