NEET · Physics · STD 11 - 4.1 newtons laws of motion
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively :
- A \(20 m s ^{-2} ; \tan ^{-1}(4 / 3)\) with 8 N force
- B \(2 m s ^{-2} ; \tan ^{-1}(3 / 4)\) with 6 N force
- C \(2 m s ^{-2} ; \tan ^{-1}(4 / 3)\) with 8 N force
- D \(2 m s ^{-2} ; \tan ^{-1}(3 / 4)\) with 8 N force
Answer & Solution
Correct Answer
(D) \(2 m s ^{-2} ; \tan ^{-1}(3 / 4)\) with 8 N force
Step-by-step Solution
Detailed explanation
(D) \(2 m s ^{-2} ; \tan ^{-1}(3 / 4)\) with 8 N force
Given forces are \(F_1=8 N\) and \(F_2=6 N\) acting perpendicular to each other.
The magnitude of the net force is \(F=\sqrt{F_1^2+F_2^2}=\sqrt{8^2+6^2}=10 N\).
The magnitude of acceleration is \(a=\frac{F}{m}=\frac{10}{5}=2 m s ^{-2}\).
Let \(\theta\) be the angle made by the net force with the 8 N force.
\(\tan \theta=\frac{F_2}{F_1}=\frac{6}{8}=\frac{3}{4}\)
\(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with the 8 N force.
Given forces are \(F_1=8 N\) and \(F_2=6 N\) acting perpendicular to each other.
The magnitude of the net force is \(F=\sqrt{F_1^2+F_2^2}=\sqrt{8^2+6^2}=10 N\).
The magnitude of acceleration is \(a=\frac{F}{m}=\frac{10}{5}=2 m s ^{-2}\).
Let \(\theta\) be the angle made by the net force with the 8 N force.
\(\tan \theta=\frac{F_2}{F_1}=\frac{6}{8}=\frac{3}{4}\)
\(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) with the 8 N force.
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