homework and exercises - Calculating impact force for a falling object?

Good evening, I'm trying to calculate what kind of impact force a falling object would have once it hit something. This is my attempt so far:

Because $x= \frac{1}{2} at^2$, $t=\sqrt{2x/a}$
$v=at$, therefore $v=a \sqrt{2x/a}$
$E_k=\frac{1}{2} mv^2$, so $E_k=\frac{1}{2} m(2ax)=m \cdot a \cdot x$
Since $W=E_k=F_i s$, $F_i=E_k/s=(m \cdot a \cdot x)/s$

For an object weighing about as much as an apple, $0.182$ kg, falling $2.00$ m straight down and creating a dent of $0.00500$ m, this would result in:

$$F_i=(m \cdot a \cdot x)/s$$

$$F_i=(0.182 \cdot 9.81 \cdot 2.00)/0.00500=706 \, \text{N}$$

Does this make any sense? I wouldn't be surprised at all to find out I'm missing something here.

Any input would be appreciated,

thanks in advance!

Answer

If your apple falls $2m$ it's velocity is calculated using the equation you give:

$$v^2 = 2as$$

and you get $v^2 = 39.24 \space m^2s^{-2}$ (I've haven't taken the square root for reasons that will become obvious). You know the apple is slowed to rest in $0.005m$, so you just need to work out what acceleration is needed when $v^2 = 39.24$ and $s = 0.005$. A quick rearrangement of your equation gives:

$$a = \frac{v^2}{2s}$$

and plugging in $v^2 = 39.24$ and $s = 0.005$ gives $a = 3925 \space ms^{-2}$. To get the force just use Newton's equation:

$$F = ma$$

where $m$ is the mass of the apple, $0.18 kg$, and you get $F = 706.32N$. So you got the correct answer (my answer differs from yours only because I used $g = 9.81 \space ms^{-2}$).

To get a more general result substitute for $v^2$ in the second equation to get:

$$F = ma = m\frac{2gs_1}{2s_2} = mg\frac{s_1}{s_2}$$

where $s_1$ is the distance the apple falls and $s_2$ is the distance it takes to stop.