Let's solve the problem step-by-step:
1. Determine the kinetic energy of the hammer head:
- The mass of the hammer head [tex]\( m \)[/tex] is 0.8 kg.
- The velocity of the hammer head [tex]\( v \)[/tex] is 5 m/s.
The formula for kinetic energy [tex]\( KE \)[/tex] is:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
Plugging in the values:
[tex]\[
KE = \frac{1}{2} \times 0.8 \, \text{kg} \times (5 \, \text{m/s})^2
\][/tex]
[tex]\[
KE = \frac{1}{2} \times 0.8 \, \text{kg} \times 25 \, \text{m}^2/\text{s}^2
\][/tex]
[tex]\[
KE = 0.4 \, \text{kg} \times 25 \, \text{m}^2/\text{s}^2
\][/tex]
[tex]\[
KE = 10 \, \text{Joules}
\][/tex]
2. Calculate the work done on the nail:
- The work done [tex]\( W \)[/tex] by the hammer is equal to the kinetic energy of the hammer head since all the kinetic energy is transferred to the nail.
[tex]\[
W = KE = 10 \, \text{Joules}
\][/tex]
3. Determine the average force exerted on the nail:
- The distance [tex]\( d \)[/tex] over which the force is applied is 6 mm, which is 0.006 meters (since 1 mm = 0.001 m).
The formula for work done is:
[tex]\[
W = F_{\text{avg}} \times d
\][/tex]
Rearranging to solve for the average force [tex]\( F_{\text{avg}} \)[/tex]:
[tex]\[
F_{\text{avg}} = \frac{W}{d}
\][/tex]
Plugging in the values:
[tex]\[
F_{\text{avg}} = \frac{10 \, \text{Joules}}{0.006 \, \text{meters}}
\][/tex]
[tex]\[
F_{\text{avg}} = \frac{10}{0.006} \, \text{Newtons}
\][/tex]
[tex]\[
F_{\text{avg}} \approx 1666.67 \, \text{Newtons}
\][/tex]
Therefore, the average force on the nail is approximately [tex]\( 1666.67 \, \text{Newtons} \)[/tex].
