Chapter 3, Problem 6C

The "shot" used in the shot-put event is a metal ball with a mass of 7.3 kg. When thrown in Olympic competition, it is accelerated to a speed of about 14 m/s. As an approximation, let's say that the athlete exerts a constant force on the shot while throwing it and that it moves a distance of 3 m while accelerating.
(a) What is the shot's kinetic energy?
(b) Compute the force that acts on the shot.

(c) It takes about 0.5 s to accelerate the shot. Compute the power required. Convert your answer to horsepower.

To determine

(a)

The kinetic energy of the shot.

Answer to Problem 6C

The kinetic energy of the shot is $715.4J$.

Explanation of Solution

Given:

The “shot” used in the shot-put event is a metal ball with a mass of $7.3kg$. When thrown in Olympic competition, it is accelerated to a speed of about $14m/s$. As an approximation, let’s say that the athlete exerts a constant force on the shot while throwing it and that it moves a distance of $3m$ while accelerating.

Formula used:

Kinetic energy is given as

$KE=\frac{1}{2}m{v}^2$

Where, $m=$ Mass of body.

$v=$ Speed of body.

Calculation:

We have,

Mass of shot, $m=7.3kg$

Speed of shot, $v=14m/s$

Thus, kinetic energy of shot is calculated as

$KE=\frac{1}{2}\left(7.3kg\right){\left(14m/s\right)}^2$

$KE=715.4J$.

Conclusion:

Hence, the kinetic energy of the shot is $715.4J$.

To determine

(b)

The force that acts on the shot.

Answer to Problem 6C

The force that acts on the shot is $238.5N$.

Explanation of Solution

Given:

The “shot” used in the shot-put event is a metal ball with a mass of $7.3kg$. When thrown in Olympic competition, it is accelerated to a speed of about $14m/s$. As an approximation, let’s say that the athlete exerts a constant force on the shot while throwing it and that it moves a distance of $3m$ while accelerating.

Formula used:

There the four equations of motions, listed below:

$d=ut+\frac{1}{2}a{t}^2$

$v=u+at$

$v^2=u^2+2ad$

$d=\frac{u+t}{2t}$

Where, $d=$ Displacement.

$u=$ Initial velocity.

$v=$ Final velocity.

$a=$ Acceleration.

$t=$ Time.

Calculation:

Given that the shot is accelerated to a speed of about $14m/s$. Also, shot moves a distance of $3m$ while accelerating That means, the initial velocity of car is zero.

Thus, we have

$u=0$

$v=14m/s$

$d=3m$

Using equation of motion, we have

$v^2=u^2+2ad$

Substituting the values, we get

${\left(14m/s\right)}^2=0^2+2a\left(3m\right)$

${\left(14m/s\right)}^2=2a\left(3m\right)$

$a=\frac{{\left(14m/s\right)}^2}{2\left(3m\right)}$

$a=\frac{196}{6}m/s^2$

$a=32.67m/s^2$

Also, we know that force is calculated as

$Force=Mass\times Acceleration$

Substituting the values, we get

$F=7.3kg\times 32.67m/s^2$

$F=238.49N$.

Conclusion:

Hence, the force that acts on the shot is $238.5N$.

To determine

(c)

The “shot” used in the shot-put event is a metal ball with a mass of $7.3kg$. When thrown in Olympic competition, it is accelerated to a speed of about $14m/s$. As an approximation, let’s say that the athlete exerts a constant force on the shot while throwing it and that it moves a distance of $3m$ while accelerating. It takes about $0.5s$ to accelerate the shot. Find the power required in horsepower.

Answer to Problem 6C

The power required is $2.24hp$.

Explanation of Solution

Given:

Given that shot used shot-put event is a metal ball with a mass of $7.3kg$ and it is accelerated to a speed of about $14m/s$. Also, it moves a distance of $3m$ while accelerating.

Formula used:

There the four equations of motions, listed below:

$d=ut+\frac{1}{2}a{t}^2$

$v=u+at$

$v^2=u^2+2ad$

$d=\frac{u+t}{2t}$

Where $d=$ Displacement.

$u=$ Initial velocity.

$v=$ Final velocity.

$a=$ Acceleration.

$t=$ Time.

Also, Power is given as,

Calculation:

Given that the shot is accelerated to a speed of about $14m/s$. Also, shot moves a distance of $3m$ while accelerating That means, the initial velocity of car is zero.

Thus, we have

$u=0$

$v=14m/s$

$d=3m$

Using equation of motion, we have

$v^2=u^2+2ad$

Substituting the values, we get

${\left(14m/s\right)}^2=0^2+2a\left(3m\right)$

${\left(14m/s\right)}^2=2a\left(3m\right)$

$a=\frac{{\left(14m/s\right)}^2}{2\left(3m\right)}$

$a=\frac{196}{6}m/s^2$

$a=32.67m/s^2$

Also, we know that force is calculated as

$Force=Mass\times Acceleration$

Substituting the values, we get

$F=7.3kg\times 32.67m/s^2$

$F=238.49N$.

Conclusion:

Hence, the power required is $2.24hp$.