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

4 sec

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=(W)/(F_a) \\ \\ d=(4900)/(490) \\ \\ d=10\text{ m} \end{gathered}`

Now, we calculate the time using the speed formula:

`\begin{gathered} v=(d)/(t) \\ \\ t=(d)/(v) \\ \\ \text{ we replacing:} \\ \\ t=(10)/(2.5)=4\text{ sec} \end{gathered}`

The time of the elevator trip is 4 seconds