Chapter 9.1, Problem 5PP

(a)

To determine

The average force exerted on the person.

Answer to Problem 5PP

  $F=7.83\times {10}^3\text{ N}$ , acting in the direction opposite to the direction of motion of the person.

Explanation of Solution

Given:

Mass of the person is $m=60\text{ kg}$

Initial velocity of the car (and that of the person) is ${\text{v}}_i=94\text{ km/h = }\frac{94\times 1000\text{ m}}{3600\text{ s}}=26.11\text{ m/s}$

Final velocity of the car (and that of the person) is ${\text{v}}_f=0\text{ m/s}$

Time interval at which velocity of the car undergoes change is $\Delta \text{t = 0}\text{.20 s}$

Formula used:

In order to find the average force exerted on the person let us use the relationship between the impulse and change in momentum. Impulse experienced by an object must be the change in momentum of that object. This is given by the expression,

  $F\Delta t=p_f-p_i$$\left(1\right)$

Where,

  $F$ is force on the object

  $\Delta t$ is the time interval

  $p_f$ is the final momentum of an object, $p_f=m{v}_f$

  $p_i$ is the initial momentum of an object, $p_i=m{v}_i$

  $m$ is mass of an object,

  $v_i$ and $v_f$ are the initial and final velocity of an object

Substituting for $p_f$ and $p_i$ in equation $\left(1\right)$ ,

  $F\Delta t=m{v}_f-m{v}_i$

  $\Rightarrow F=\frac{m{v}_f-m{v}_i}{\Delta t}$$\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $F=\frac{\left(60\text{ kg}\right)\left(0\text{ m/s}\right)-\left(60\text{ kg}\right)\left(26.11\text{ m/s}\right)}{\text{0}\text{.20 s}}$

  $\begin{array}{l}=\frac{0-1566.6\text{ kg}\cdot \text{m/s}}{0.20\text{ s}}\\ =\text{ -7833 kg}\cdot {\text{m/s}}^2\\ =-7.83\times {10}^3\text{ N}\end{array}$

OR

  $F=7.83\times {10}^3\text{ N}$ , acting in the direction opposite to the direction of motion of the person.

Conclusion:

The average force exerted on the person is $7.83\times {10}^3\text{ N}$ acting in the direction opposite to the direction of motion of the person.

(b)

To determine

The mass of an object that has a weight equal to the force calculated in the part (a), whether this mass can be able to lift that weight by a person, and whether a person is strong enough to stop his body with his arms.

Answer to Problem 5PP

  $m=799\text{ kg}$ . This much mass is too heavy to lift. A person cannot safely stop his body with his arms

Explanation of Solution

Given:

Mass of the person is $m=60\text{ kg}$

The average force $\left(F\right)$ exerted on the person is $7.83\times {10}^3\text{ N}$ or $7.83\times {10}^3\text{ kg}\cdot {\text{m/s}}^2$

Formula used:

Force $\left(F\right)$ and mass $\left(m\right)$ is connected by Newton’s second law of motion,

  $F=ma$$\left(3\right)$

For weight calculation, acceleration $\left(a\right)$ is can be replaced by acceleration due to gravity $\left(g\right)$ , $g=9.8{\text{ m/s}}^2$

Hence the equation can be written as,

  $F=mg$

  $\Rightarrow m=\frac{F}{g}$$\left(4\right)$

Calculation:

Substituting the numerical values in equation $\left(4\right)$

  $\Rightarrow m=\frac{7.83\times {10}^3\text{ kg}\cdot {\text{m/s}}^2}{9.8{\text{ m/s}}^2}$

  $=799\text{ kg}$

Conclusion:

The mass of an object that has a weight equal to the force calculated in the part (a) is $799\text{ kg}$ . This much mass is too heavy to lift. A person cannot safely stop his body with his arms