Chapter 4, Problem 4P

According to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

To determine

The magnitude of the force exerted by the heart muscle.

Answer to Problem 4P

Solution:

The force exerts by heart muscle is $0.02\text{ N}$.

Explanation of Solution

The concept we must use to solve this problem is average acceleration and force. The average acceleration is the change in velocity with respect to the elapsed time. So, the expression for the average acceleration is given as below:

$a_{\text{avg}}=\frac{v_f-v_i}{t_f-t_i}$ …… ……(1)

Here, $v_f$ is the final velocity, $v_i$ is the initial velocity, $t_f$ is the final time and $t_i$ is the initial time.

The force on an object is calculated by using Newton’s second law of motion as given below:

$F_{\text{avg}}=m{a}_{\text{avg}}$………… (2)

Given:

Mass of the blood $m=20\text{ g}$

Initial time $t_i=0\text{ s}$

Initial speed $v_i=0.25\text{ m/s}$

Final speed $v_f=0.35\text{ m/s}$

Final Time $t_f=0.10\text{ s}$

Formula used:

$a_{\text{avg}}=\frac{v_f-v_i}{t_f-t_i}$

$F_{\text{avg}}=m{a}_{\text{avg}}$

Calculation:

Substitute, $v_i=0.25\text{ m/s}$, $v_f=0.35\text{ m/s}$ and $t_f=0.10\text{ s}$, $t_i=0\text{ s}$ in the equation (1)so the average acceleration is given as below:

${\text{a}}_{\text{avg}}=\frac{v_f-v_i}{t_f-t_i}$

       $=\frac{0.25\text{ m/s}-0.35\text{ m/s}}{0.10\text{ s}-0}$

       =1 m/s²

Now we can find the magnitude of force exerted by heart muscle by substituting $m=20\text{ g}$ and $a_{\text{avg}}=1{\text{ m/s}}^2$ in equation (2)as given below:

F_avg=ma_avg

      =(20 g)(1m/s²)

      =0.02N