Two forces, one of 3 newton and another of 4 newton are applied on a s

Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below:
The magnitude of the resultant acceleration is:

7 m/s²

1 m/s²

5 m/s²

√7 m/s²

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-2 – 2015

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The correct option is C, 5 m/s².

Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $\sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ N. According to Newton’s second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N / 1 kg = 5 m/s².

The direction of the resultant force (and thus acceleration) would be at an angle to the x-axis, given by θ = atan(Fy/Fx) = atan(4/3). However, the question only asks for the magnitude of the acceleration. This problem illustrates how to find the resultant of perpendicular forces and apply Newton’s second law.

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