Question

If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator maybe ________.

• A. going up with the increasing speed
• B. going down with increasing speed
• C. going up with the uniform speed
• D. going down with uniform speed

Answer 1

The correct option is D

going down with uniform speed

Explanation for the correct Answer:

Given, tension ($T$) in the cable supporting the elevator is equal to weight of the elevator($W$).

The formula for tension in a string is,
$T=W+ma$
where $m$ is the mass of the elevator connected to the string and $a$ is the acceleration of the elevator.

In case of option C,

The free body diagram is as follows,

The elevator is moving upward at a uniform speed which means that the acceleration of the elevator, $a=0$. Thus,
$\begin{array}{rcl}T& =& W+m·0\\ & \Rightarrow & T=W\end{array}$

So, option C is correct.

In case of option D,

The free body diagram is as follows,

The elevator is moving downward at a uniform speed which means that the acceleration of the elevator, $a=0$. Thus,
$\begin{array}{rcl}T& =& W+m·0\\ & \Rightarrow & T=W\end{array}$

So, option D is correct.

Explanation for Incorrect Answer:

In case of option A,

The free body diagram is as follows,

The elevator is moving upward at an increasing speed which means that the acceleration of the elevator, $a>0$. Thus,

$\begin{array}{rcl}T& =& W+m·a\\ & \Rightarrow & T\ne W\end{array}$

Hence, option A is incorrect.

In case of option B,

The free body diagram is as follows,

The elevator is moving downward at an increasing speed which means that the acceleration of the elevator, $a<0$ (It is negative because the direction of acceleration is opposite to the previous case). Thus,

$\begin{array}{rcl}T& =& W-m·a\\ & \Rightarrow & T\ne W\end{array}$

Hence, option D is also incorrect.

Hence, the correct options are C and D.