Sensitivity (S) relates the output signal to the input measurand
Here, sensitivity relates output voltage (V) to input force (F)
$$ S = \frac{\text{Output Voltage}}{\text{Input Force}} = \frac{V}{F} $$
We are given:
Sensitivity \(S = 0
01\) mV/N = \(0
01 \times 10^{-3}\) V/N
Output Voltage \(V = 2\) mV = \(2 \times 10^{-3}\) V
We need to find the force \(F\)
Rearranging the formula:
$$ F = \frac{V}{S} $$
$$ F = \frac{2 \times 10^{-3} \, \text{V}}{0
01 \times 10^{-3} \, \text{V/N}} $$
$$ F = \frac{2}{0
01} \, \text{N} = \frac{200}{1} \, \text{N} = 200 \, \text{N} $$
The corresponding force is 200 N