Question

A body moves with velocity $ v = \ln x\dfrac{m}{s} $ where x is its position. The net force acting on body is zero at:
(A) 0 m
(B) $ x = {e^2}m $
(C) $ x = em $
(D) $ x = 1m $

Answer 1

Hint :In order to solve this question, you must be aware about the relationship between the velocity of an object and its position. If position is given by a function p(x), then the velocity is the first derivative of that function, and the acceleration is the second derivative.

Complete Step By Step Answer:
Given: Velocity of body = $ \ln (x) $
To find: The net force acting on the body is zero at x = ?
We know that, velocity v of a body is given by $ v = \dfrac{{ds}}{{dt}} $
Where, s is the displacement of the body at time t
Acceleration of the body is given by $ a = \dfrac{{dv}}{{dt}} $
Where, v is the velocity of the body at time t
 $ \dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x} $
Let the mass of the body be m and the acceleration of the body be a
Then, force acting on the object is given as the product of its mass and its acceleration.
 $ F = ma $
 $ a = \dfrac{{dv}}{{dt}} $
Multiplying and dividing the R.H.S. by dx, we get:
 $ a = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}} $
 $ a = \dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}} $
 $ \dfrac{{dx}}{{dt}} $ is the velocity of the given body
Therefore, $ a = \dfrac{d}{{dx}}[\ln (x)] \times v $
 $ a = \dfrac{1}{x} \times \ln (x) $
 $ a = \dfrac{{\ln (x)}}{x} $
Let the net force acting on the given body be F
 $ F = ma $
 $ F = m\dfrac{{\ln (x)}}{x} $
It is given that the net force acting on the object is 0
Therefore, $ 0 = m\dfrac{{\ln (x)}}{x} $
 $ m\dfrac{{\ln (x)}}{x} = 0 $
 $ \dfrac{{ln(x)}}{x} = 0 $
 $ \ln (x) = 0 $
 $ x = {e^0} $
 $ x = 1 $
The net force acting on the body is zero at $ x = 1 $
Hence, Option (D) is correct.

Note :
If the net force acting on a body is zero, then the body won’t necessarily be in the rest position. When the net force of the body is equal to zero, it implies that the body cannot be accelerated. This follows from Newton’s second law of motion.