According to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from$\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{m}\mathbf{/}\mathbf{s}$to$\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{m}\mathbf{/}\mathbf{s}$during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

According to this law, the heart muscle’s force depends on the blood contained in each pulse and the acceleration of the blood.

The product of each pulse’s mass and blood acceleration can be used to estimate the force exerted by the heart to accelerate blood.

The given data can be listed below as:

• The mass of each pulse of the blood is $m=20\mathrm{g}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}=0.02\mathrm{kg}$.
• The period of acceleration of the blood is $t=0.10\mathrm{s}$.
• The initial velocity of the blood is, $v_i=0.25\mathrm{m}/\mathrm{s}$.
• The final velocity of the blood is $v_f=0.35\mathrm{m}/\mathrm{s}$.

From Newton’s first law, the final velocity of the blood can be expressed as,

$\begin{array}{c}{v}_f=v_i+at\\ a=\frac{v_f-v_i}{t}\end{array}$

Here, a is the acceleration of the blood.

Substitute the values as $0.35\mathrm{m}/\mathrm{s}$ for $v_f$, $0.25\mathrm{m}/\mathrm{s}$ for $v_i$, and $0.10\mathrm{s}$ for t in the above equation.

$\begin{array}{c}a=\frac{0.35\mathrm{m}/\mathrm{s}-0.25\mathrm{m}/\mathrm{s}}{0.10\mathrm{s}}\\ =1\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the acceleration of the blood is $1\mathrm{m}/{\mathrm{s}}^2$.

From Newton’s second law, the force exerted by the heart muscle can be given as:

$F=ma$

Here, m is the mass or quantity of the blood, and a is the acceleration of the blood.

Substitute the values as 0.02 kg for m, and $1\mathrm{m}/{\mathrm{s}}^2$for a in the above equation.

$\begin{array}{c}F=0.02\mathrm{kg}\times 1\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =0.02\mathrm{N}\end{array}$

Thus, the force exerted by the heart muscle is 0.02 N.