91Ó°ÊÓ

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,

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

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,

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

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,

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

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$98W$.