Question

a person strikes a nail with a hammer of mass 500 g moving with a velocity of 10 m/s . the hammer comes to rest in 0.01 s after striking the nail . calculate the force exerted by hammer on nail. and the distance moved by the nail in to the plank.

Answer 1

Given,
Mass of hammer, m=500 g=0.5 kg
Initial velocity of hammer, u=10 m/s
Final velocity of hammer, v=0
Duration=0.01 a
Using second law of motion,
F=ma
=(v-u) /t
=0.5 (0-10)/0.01=-500 N
The negative sign shows the force of 500 N acts in the opposite direction of motion.
Thus, force exerted by hammer on nail=500 N
Again,
$$\begin{gathered} {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as} \\ 0-100=2\times \frac{(0-10)}{0.01}\times \mathrm{s} \\ \mathrm{s}=\frac{100\times 0.01}{2\times 10}=0.05\mathrm{m} \end{gathered}$$