Question

A block of mass m is dropped from the fourth floor of an office building and hits the sidewalk below at speed V. From what floor should the block be dropped to double that impact speed? (a) the sixth floor (b) the eighth floor (c) the tenth floor (d) the twelfth floor (e) the sixteenth floor.

Short answer

The correct option is (e): the sixteenth floor.

Step-by-step solution

Given information

The mass of the block is m.

The block is dropped from the fourth floor.

The final speed of the block is V.

Mechanical energy

The mechanical energy of a body can be determined by adding the ‘total kinetic’ and ‘gravitational potential’ energy of the body.

For a body free falling from a certain height, there is only two energies associated with the body: kinetic and potential energy.

The floor height for double impact speed

When the block falls freely from a certain height, only the conservative gravitational force acts on it. Then, the mechanical energy of the system is conserved, and its expression is given as follows:

$\begin{array}{l}\Delta KE+\Delta PE=0\\ K{E}_f-K{E}_i+P{E}_f-P{E}_i=0\\ \frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2+mg{h}_f-mg{h}_i=0\end{array}$

Put the final speed of the block ${\mathrm{v}}_{\mathrm{f}}=\mathrm{v}$, the initial speed of the block ${\mathrm{v}}_{\mathrm{i}}=0$, the final height of the block ${\mathrm{h}}_{\mathrm{t}}=0$, and the initial height of the block from the fourth floor ${\mathrm{h}}_{\mathrm{i}}=\mathrm{h}$.

$\begin{array}{l}\frac{1}{2}m{v}^2-0+0-mgh=0\\ \frac{1}{2}m{v}^2=mgh\\ h=\frac{v^2}{2g}\\ 4h=\frac{{\left(2v\right)}^2}{2g}\end{array}$

From the above expression, the final speed of the block can be doubled by increasing the initial height by a factor of four.

So, the floor height to double the impact speed of the block can be calculated as follows:

$\begin{array}{l}\mathrm{Height}\mathrm{for}\mathrm{double}\mathrm{impact}\mathrm{speed}=4\times \left(4\mathrm{th}\mathrm{Floor}\right)\\ \mathrm{Height}\mathrm{for}\mathrm{double}\mathrm{impact}\mathrm{speed}=16\mathrm{th}\mathrm{Floor}\end{array}$

Hence, the block should be dropped from the 16th floor to double the impact speed, and the correct option is (e), the sixteenth floor.