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 accelera

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Q: 49 (NDA-II/2015)
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:

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,3,3,1,2,3,0

keywords: 

{'resultant acceleration': [0, 0, 0, 1], 'forces': [0, 0, 0, 1], 'frictionless surface': [0, 1, 0, 2], 'newton': [0, 1, 0, 0], 'axis': [0, 0, 0, 1], 'magnitude': [0, 0, 3, 10], 'kg body': [0, 0, 0, 1], 's2': [0, 1, 2, 6]}

In this question, two forces are applied to a 1 kg body. One force of 3 N is applied along the x-axis and another force of 4 N is applied along the y-axis. We want to find the magnitude of the resultant acceleration.

To find the resultant acceleration, we can use Newton`s second law of motion, which states that F = ma, where F is the net force applied, m is the mass of the object, and a is the acceleration.

Here, we can calculate the net force applied along the x-axis by finding the x-component of the force applied along the y-axis. Since these forces are perpendicular to each other, their components are also perpendicular. So, the net force in the x-direction is 3 N.

Similarly, the net force applied along the y-axis can be calculated by finding the y-component of the force applied along the x-axis. The y-component of the force is 4 N.

Now, we can use Pythagoras theorem to find the magnitude of the resultant force. The magnitude of the resultant force is the square root of the sum of the squares of the x-component and the y-component of the forces.

Using the Pythagorean theorem, we have:

Resultant force = √(3^