Question

A 50 kg person is taking a ride on an elevator travelling up at a steady speed of 2.5 m/s. Find the time of the elevator trip if the elevator does 4900 J of work on the person.

Answer 1

ANSWER:

4 sec

STEP-BY-STEP EXPLANATION:

Given:

Mass (m) = 50 kg

Speed (v) = 2.5 m/s

Work (W) = 4900 J

Speed is constant then a = 0

Net applying force = 0 N

We make the vertical force balance:

$\begin{gathered} F_{net}=F_a-F_g \\ \\ 0=F_a-F_g\rightarrow F_a=F_g \\ \\ F_g=m\cdot g \\ \\ \text{ We replacing} \\ \\ F_g=50\cdot9.8 \\ \\ F_g=490\text{ N} \\ \\ F_a=490\text{ N} \end{gathered}$

Now we calculate the distance:

$\begin{gathered} W=F_a\cdot d \\ \\ d=\frac{W}{F_a} \\ \\ d=\frac{4900}{490} \\ \\ d=10\text{ m} \end{gathered}$

Now, we calculate the time using the speed formula:

$\begin{gathered} v=\frac{d}{t} \\ \\ t=\frac{d}{v} \\ \\ \text{ we replacing:} \\ \\ t=\frac{10}{2.5}=4\text{ sec} \end{gathered}$

The time of the elevator trip is 4 seconds