An elevator in a tall building is allowed to reach a maximum speed of$3.5\mathrm{m}/\mathrm{s}$going down. What must the tension be in the cable to stop this elevator over a distance of 2.6 m if the elevator has a mass of 1450 kg including occupants?

When the direction of acceleration is opposite to the direction of initial velocity, the velocity of the body reduces over time. This type of motion is called retarding motion.

The mass of the lift, including occupants,$M=1450\mathrm{kg}$

The maximum speed of the lift in the downward direction,$v=3.5\mathrm{m}/\mathrm{s}$

Let the acceleration of the lift in the upward direction be a.

As the lift must stop over a distance of 2.6 m, using the third equation of motion, you can evaluate:

$0-v^2=-2a\times 2.6$

Substituting the value of v, you get:

$3.5^2=2a\times 2.6$

Solving the above equation for a, you get:

$a=2.36\mathrm{m}/{\mathrm{s}}^2$

The FBD of the system can be drawn as:

T is the tension force in the cable.

Applying Newton’s second law on the system, you get:

$T-Mg=Ma$

Substituting the values in the above equation, you get:

$\begin{array}{c}T=1450\left(9.8+2.36\right)\\ =17632\mathrm{N}\end{array}$

Thus, the tension force in the cable is 17632 N.