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The magnitude and direction of the acceleration produced in a body of mass $5 \text{ kg}$ when two mutually perpendicular forces $8 \text{ N}$ and $6 \text{ N}$ act on it, are respectively :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$2 \text{ m s}^{-2}, \tan^{-1}(3/4)$ with $8 \text{ N}$ force

Calculate Acceleration Magnitude and Direction for Perpendicular Forces

This solution explains how to find the acceleration's magnitude and direction when forces act perpendicularly on an object.

Given Information

  • Mass of the body, $m = 5 \text{ kg}$
  • First force, $F_1 = 8 \text{ N}$
  • Second force, $F_2 = 6 \text{ N}$
  • The forces $F_1$ and $F_2$ are mutually perpendicular.

Calculate Net Force

Since the forces are perpendicular, we find the magnitude of the resultant net force ($F_{net}$) using the Pythagorean theorem:

$ F_{net} = \sqrt{F_1^2 + F_2^2} $

Substituting the values:

$ F_{net} = \sqrt{(8 \text{ N})^2 + (6 \text{ N})^2} $

$ F_{net} = \sqrt{64 \text{ N}^2 + 36 \text{ N}^2} $

$ F_{net} = \sqrt{100 \text{ N}^2} $

$ F_{net} = 10 \text{ N} $

Calculate Acceleration Magnitude

Using Newton's second law, $F_{net} = m \times a$, we can find the acceleration ($a$):

$ a = \frac{F_{net}}{m} $

Substituting the values:

$ a = \frac{10 \text{ N}}{5 \text{ kg}} $

$ a = 2 \text{ m s}^{-2} $

Determine Acceleration Direction

The acceleration is in the direction of the net force. We find the angle $\theta$ the net force makes with the $8 \text{ N}$ force.

The tangent of this angle is the ratio of the perpendicular force ($F_2$) to the force it's measured against ($F_1$):

$ \tan(\theta) = \frac{F_2}{F_1} $

$ \tan(\theta) = \frac{6 \text{ N}}{8 \text{ N}} $

$ \tan(\theta) = \frac{3}{4} $

Therefore, the angle is:

$ \theta = \tan^{-1}\left(\frac{3}{4}\right) $

The direction is $\tan^{-1}(3/4)$ with the $8 \text{ N}$ force.

Final Result

The magnitude of the acceleration is $2 \text{ m s}^{-2}$ and its direction is $\tan^{-1}(3/4)$ with the $8 \text{ N}$ force.

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