Answered

If the 0.8 kg hammer's head is moving at 5 m/s when it strikes a nail 6 mm into a piece of wood, what is the average force on the nail?



Answer :

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].