Question

An elevator is descending an elevator shaft, and uniformly comes to rest over a distance of 30 m. If the elevator’s initial speed was 8 m/s, and its mass is 500 kg, what net force did the elevator experience?

Answer 1

Given:

Initial speed = 8 m/s

Distance covered = 30 m

Mass of elevator = 500 kg

Let's find the net force the elevator experienced.

,Apply the formula below to find the acceleration:

`v^2=u^2+2as`

Where:

v is the fainal velocity = 0 m/s

u is the initial velocity = 8 m/s

s = 30 m

Thus, to solve for the acceleration, a, we have:

`\begin{gathered} a=(v^2-u^2)/(2s) \\ \\ a=(0^2-8^2)/(2*30) \\ \\ a=(-64)/(60) \\ \\ a=-1.067m/s^2 \end{gathered}`

The acceleration of the elevator is -1.067 m/s².

Now, to find the Net force, apply the formula:

`F=ma`

Where:

F is the net force

m is the mass = 500 kg

a = -1.067 m/s².

We have:

`\begin{gathered} F=500*(-1.067) \\ \\ N=-533.33\text{ N} \end{gathered}`

Therefore, the net force is 533.33 N upward.

ANSWER:

533.33 N upward.