Question

A hammer of mass $500\mathrm{g}$ moving at $50\mathrm{m}/\mathrm{s}$, strikes a nail. The nail stops the hammer in a very short time of $0.01\mathrm{s}$. What is the force of the nail on the hammer?

Answer 1

Step 1: Given data

Mass of the hammer (m) is $500g=0.5\mathrm{kg}$

The initial velocity of the hammer is $\mathrm{u}=50\mathrm{m}/\mathrm{s}$

The nail takes time to stop the hammer $t=0.01\mathrm{s}$

The final velocity of the hammer $\mathrm{v}=0$

Step 2: Formula used

According to Newton's second law, the force of the nail on the hammer is as follows:

$f=\frac{m(v–u)}{t}$

Where,

$f$is the force.

$v$ is the final velocity.

$u$ is the initial velocity.

$t$ is the time.

Step 3: Finding the force of the nail on the hammer.

By substituting the value of $v$, $u$ and $t$ in the above-mentioned formula to find the force of the nail on the hammer:

$\begin{array}{c}f=\frac{0.5\times \left(0-50\right)}{0.01}\\ =-\frac{0.5\times 50}{0.01}\\ =-2500 \mathrm{N}\end{array}$

$-\mathrm{ve}$ the sign indicates the force acting by the nail on the hammer is opposite to the direction of the motion of the hammer.

Hence, the magnitude of the force of the nail on the hammer is $2500\mathrm{N}$.