Newton´s second law and the nail in the wall

Data

• Mass of the hammer m hammer= 2 kg
• Initial hammer´s velocity v_o = 6 m/s
• Traveled distance by the nail x = 5 mm = 5·10^-3 m

Previous considerations

We can suppose the hammer and the nail as one single element drilling in the wood and, progressively, deacreasing its velocity due to the force opposite to the wood. We will guess this force the wood puts up is constant, and the mass of the nail is insignificant opposite the hammer´s.

Resolution

The strategy to solve the problem consist of finding the acceleration the hammer - nail ensemble is subdued , and from that and Newton´s second law, to calculate the force of resistance of the wall.

The ensemble hammer - nail has an initial velocity of v_o = 6 m/s, and a final one of v_f = 0, since the force the wood puts up makes it slow down. By assuming the force is constant, the acceleration will also be it, so we find a u.a.r.m whose equations are (if we suppose we go in the axis x):

$$\begin{gathered} a=cte \\ {v}_f=v_0+a·t \\ x=x_0+v_0·t+\frac{1}{2}·a·t^2 \end{gathered}$$

From the second and the third equation we can obtain the acceleration the hammer - nail system is subdued. For that we propose the following process:

$$\left.\begin{array}{r}{v}_f=v_0+a·t\\ x=x_0+v_0·t+\frac{1}{2}·a·t^2\end{array}\right\}\left.\begin{array}{r}0=6+a·t\\ 5·{10}^{-3}=0+6·t+\frac{1}{2}·a·t^2\end{array}\right\}\left.\begin{array}{r}a·t=-6\\ 5·{10}^{-3}=0+6·t+\frac{1}{2}·a·t·t\end{array}\right\}$$

And we replace the upper equation for the lower one:

$$5·{10}^{-3}=0+6·t+\frac{1}{2}·a·t·t\Rightarrow 5·{10}^{-3}=6·t+\frac{1}{2}·\left(-6\right)·t\Rightarrow 5·{10}^{-3}=3·t\Rightarrow t=\frac{5}{3}·{10}^{-3}s$$

Replacing the firs equation...

$$0=6+a·t\Rightarrow -6=a·\frac{5}{3}·{10}^{-3}\Rightarrow a=-\frac{18}{5}·{10}^{-3} m/s^2$$

Where the sign - notes that the velocity is decreasing, as we can expect from a resistance force. On the other hand, observe that, equations presented are all the ones you need to resolve any problem of u.a.r.m. though. We could have used the following equation to get to the same result in a more immediate way:

$$v^2=v_0^2+2·a·∆x$$

Observe that you can get that last equation through the previous ones such as is presented on the chapter of equations of u.a.r.m.

In any case, once we have obtained the acceleration, we can calculate the force requested based on Newton´s second law:$$F=m·a\Rightarrow F=2·\left(-\frac{18}{5}·{10}^{-3}\right)=7.2·{10}^{-3}\hspace{0.17em}N$$