A tense ride down

Relevant wiki: Newton's Laws of Motion

By applying Newton's second law we get $\sum F= ma$ Now $md= T + W$ where $T$ is the tension of the cable, $W$ is the weight of the elevator car, and $d$ is the deceleration of the elevator.

Using $v_f^2= v_i^2+2ds$ to get the acceleration we find $d= -\frac{\left(\SI{10}{\meter\per\second}\right)^2}{2\times\SI{25}{\meter}}=-\SI{2}{\meter\per\second\squared}$

Now, using Newton's law T=m\left(g+d\right) = \SI{800}{\kilo\gram}\times\SI{\left(9.8+2\right)}{\meter\per\second\squared}=\SI{9440}{\newton}