91Ó°ÊÓ

The given data is listed below as:

• The weight of the nail is,$\stackrel{â‡\P Ä}{\mathrm{w}}=4.9\mathrm{N}$
• The initial downward velocity of the hammer head is,$\mathrm{u}=3.2\mathrm{m}/\mathrm{s}$
• The distance moved by the hammer head is,$\mathrm{s}=0.45\mathrm{cm}×\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)=0.45×{10}^{-2}\mathrm{m}$
• The weight of the person is,${\stackrel{→}{F}}_{person}=15N$
• The distance travelled by the hammer head before coming to rest is,$d=0.12cm×\left(\frac{{10}^{-2}m}{1cm}\right)=0.12×{10}^{-2}m$

The kinematic equation mainly describes the motion of a particle which moves under constant acceleration. The square of the final velocity of the particle is equal to the addition of the square of the initial velocity and twice of the product of the acceleration and the distance moved by the particle.

The free body diagram of the hammer head has been provided below:

In the above diagram, the force exerted by the person${\stackrel{→}{F}}_{person}$ is directed downwards along with the weight of the hammer head $\stackrel{→}{w}$. Moreover, the force exerted by the nail${\stackrel{→}{F}}_{nail}$ is exerted upwards.

The action force is the force exerted by the hammer in the above diagram. The action force subsequently produces a reaction force which is the force exerted by the nail ${\stackrel{→}{F}}_{nail}$.

Thus, the reaction force is the force exerted by the nail.

The equation of the acceleration of the hammerhead is expressed as:

$v^2=u^2+2as$

Here, v is the final speed and u is the initial speed of the hammer head. s is thedistance moved by the hammer head and a is the acceleration of the hammer head.

As the hammer head finally came to rest, hence the final velocity of the hammer head is zero.

Substitute 0 for v in the above equation.

$$\begin{gathered} 0=u^2+2as \\ a=\frac{-u^{2}s}{2s} \end{gathered}$$

The equation of the downward force exerted by the nail is expressed as:

${\stackrel{→}{F}}_{nail}=w+{\stackrel{→}{F}}_{person}-\frac{w}{g}a$

Here,${\stackrel{→}{F}}_{nail}$ is the force exerted by the nail, w is the weight of the nail,role="math" localid="1665056884982" ${\stackrel{→}{F}}_{person}$ is the weight of the person, g is the acceleration due to gravity and a is the acceleration of the hammer.

Substitute the value of the equation (i) in the above equation.

$$\begin{gathered} {\stackrel{→}{F}}_{nail}=w+{\stackrel{→}{F}}_{person}-\frac{w}{g}\left(\frac{-u^2}{2s}\right) \\ =w+{\stackrel{→}{F}}_{person}+\frac{w}{g}\left(\frac{u^2}{2s}\right) \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{nail}}=4.9\mathrm{N}+15\mathrm{N}+\frac{4.9\mathrm{N}}{9.81\mathrm{m}/{\mathrm{s}}^2}\left(\frac{{\left(3.2\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.45×{10}^{-2}\mathrm{m}\right)}\right) \\ =19.9\mathrm{N}+0.49\mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\left(\frac{{\left(10.24{\mathrm{m}}^2/\mathrm{s}\right)}^2}{\left(9×{10}^{-3}\mathrm{m}\right)}\right) \\ =19.9\mathrm{N}+\left(0.49\mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\right)×\left(1137.7\mathrm{m}/{\mathrm{s}}^2\right) \\ =19.9\mathrm{N}+557.5\mathrm{N} \end{gathered}$$

Hence, further as:

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{nail}}=19.9\mathrm{N}+557.5\mathrm{N} \\ =577.4\mathrm{N} \end{gathered}$$

As the magnitude of the force exerted by the nail and the force exerted by the hammer head is equal, then the force exerted by the hammer head is 577.4 N.

Thus, the downward force exerted by the hammer head is 557.4 N.

The equation of the acceleration of the hammer head is expressed as:

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

Here, v is the final speed and u is the initial speed of the hammer head. d is thedistance travelled by the hammer head before coming to rest and a is the acceleration of the hammer head.

As the hammer head finally came to rest, hence the final velocity of the hammer head is zero.

Substitute 0 for v in the above equation.

$$\begin{gathered} 0+u^2+2ad \\ a=\frac{-u^2}{2d} \end{gathered}$$

The equation of the downward force exerted by the nail is expressed as:

${\stackrel{→}{F}}_{nail}=w+{\stackrel{→}{F}}_{person}-\frac{w}{g}a$

Here,${\stackrel{→}{F}}_{nail}$ is the force exerted by the nail, w is the weight of the nail,${\stackrel{→}{F}}_{person}$ is the weight of the person, g is the acceleration due to gravity and is the acceleration of the hammer.

Substitute the value of the equation (i) in the above equation.

$$\begin{gathered} {\stackrel{→}{F}}_{nail}=w+{\stackrel{→}{F}}_{person}-\frac{w}{g}\left(\frac{-u^2}{2d}\right) \\ =w+{\stackrel{→}{F}}_{person}+\frac{w}{g}\left(\frac{-u^2}{2d}\right) \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{nail}}=4.9\mathrm{N}+15\mathrm{N}+\frac{4.9\mathrm{N}}{9.81\mathrm{m}/{\mathrm{s}}^2}\left(\frac{{\left(3.2\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.12×{10}^{-2}\mathrm{m}\right)}\right) \\ =19.9\mathrm{N}+0.49\mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\left(\frac{\left(10.24{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\left(2.4×{10}^{-3}\mathrm{m}\right)}\right) \\ =19.9\mathrm{N}+\left(0.49\mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\right)×\left(4266.6\mathrm{m}/{\mathrm{s}}^2\right) \\ =19.9\mathrm{N}+2090.66\mathrm{N} \end{gathered}$$

Hence, further as:

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{nail}}=19.9\mathrm{N}+2090.66\mathrm{N} \\ =2110.56\mathrm{N} \end{gathered}$$

As the magnitude of the force exerted by the nail and the force exerted by the hammer head is equal, then the force exerted by the hammer head is 2110.56 N.

Thus, the force exerted by the hammer head is 2110.56 N .