91Ó°ÊÓ

Third law of motion: According to the third law of motion, every action has equal but opposite reaction force.

Suppose body A exerts a force$F_{AB}$on another body B, then according to Newton’s third law body B also exert the force$F_{AB}$on body A which is,

$F_{AB}=-F_{AB}$

(a)

According to work energy theorem,

$\begin{array}{rcl}W& =& \frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2\\ F_{g}s& =& \frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2\end{array}$

Here,$F_g$is the force exerted on the gloves,m is the mass of the arm ,v is the final velocity of the gloves ($v=0$as the gloves comes to rest after hitting opponent’s face), u is the initial velocity of the gloves$\left(u=10\text{m}/\text{s}\right)$, and s is the distance covered by the gloves before it comes to rest$\left(s=7.5\text{cm}\right)$.

The expression for the force exerted on the gloves is,

$F_g=\frac{m\left(v^2-u^2\right)}{2s}$

Putting all known values,

$\begin{array}{rcl}{F}_g& =& \frac{\left(7\text{kg}\right)×\left[{\left(0\right)}^2-{\left(10\text{m}/\text{s}\right)}^2\right]}{2×\left(7.5\text{cm}\right)}\\ & =& \frac{\left(7\text{kg}\right)×\left[{\left(0\right)}^2-{\left(10\text{m}/\text{s}\right)}^2\right]}{2×\left(7.5\text{cm}\right)×\left(\frac{1\text{m}}{100}\right)}\\ & =& -4666.67\text{N}\end{array}$

According to the Newton’s third law of motion, the force exerted by gloves on opponent’s face is,

$F_f=-F_g$

Putting all known values,

$\begin{array}{rcl}{F}_f& =& -\left(-4666.67\text{N}\right)\\ & =& 4666.67\text{N}\end{array}$

Therefore, the required force exerted by the gloves on the opponent’s face is 4666.67 N.

(b)

According to work energy theorem,

$$\begin{gathered} W\text{'}=\frac{1}{2}mv{\text{'}}^2-\frac{1}{2}mu{\text{'}}^2 \\ {F}_{k}s\text{'}=\frac{1}{2}mv{\text{'}}^2-\frac{1}{2}mu{\text{'}}^2 \end{gathered}$$

Here,is the force exerted on knuckles,m is the mass of the arm $\left(m=7\text{kg}\right)$,$v\text{'}$is the final velocity of the knuckles (as the knuckles comes to rest after hitting opponent’s face),$u\text{'}$is the initial knuckles of the gloves$\left(u\text{'}=10\text{m}/\text{s}\right)$, and$s\text{'}$is the distance covered by the knuckles before it comes to rest$\left(s\text{'}=2\text{cm}\right)$.

The expression for the force exerted on the knuckles is,

$F_k=\frac{m\left({v\text{'}}^2-{u\text{'}}^2\right)}{2s\text{'}}$

Putting all known values,

$\begin{array}{rcl}{F}_k& =& \frac{\left(7\text{kg}\right)×\left[{\left(0\right)}^2-{\left(10\text{m}/\text{s}\right)}^2\right]}{2×\left(\text{2 cm}\right)}\\ & =& \frac{\left(7\text{kg}\right)×\left[{\left(0\right)}^2-{\left(10\text{m}/\text{s}\right)}^2\right]}{2×\left(\text{2 cm}\right)×\left(\frac{1\text{m}}{100}\right)}\\ & =& -17500\text{N}\end{array}$

According to the Newton’s third law of motion, the force exerted by knuckles on opponent’s face is,

$F{\text{'}}_f=-F_k$

Putting all known values,

$\begin{array}{rcl}F{\text{'}}_f& =& -\left(-17500\text{N}\right)\\ & =& 17500\text{N}\end{array}$

Therefore, the required force exerted by the knuckles on the opponent’s face is 17500N.

(c)

When the boxing gloves are used, the force is considerably reduced than that with the knuckle punch. Even though with the gloves the force is enough to cause the damage to the opponent’s face.