Problem 370 - JEE Physics

Question

A force of \(8 \mathrm{~N}\) acts on an object of mass \(5 \mathrm{~kg}\) in \(\mathrm{X}\) -direction and another force of \(6 \mathrm{~N}\) acts on it in \(\mathrm{Y}\) -direction. Hence, the magnitude of acceleration of object will be (A) \(1.5 \mathrm{~ms}^{-2}\) (B) \(2.0 \mathrm{~ms}^{-2}\) (C) \(2.5 \mathrm{~ms}^{-2}\) (D) \(3.5 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

The magnitude of the acceleration of the object is \(2~\frac{m}{s^{2}}\).

Step 1: Find the net force in X and Y directions

To find the net force in the X and Y directions, we can represent each force as a vector with components in the X and Y directions. The given forces on the object are: \(F_1 = 8~N\) in the X-direction and \(F_2 = 6~N\) in the Y-direction Since the forces are acting in perpendicular directions, their components in the X and Y directions can be represented as: \(F_1x = 8~N\), \(F_1y = 0~N\) and \(F_2x = 0~N\), \(F_2y = 6~N\) The net force in the X direction, \(F_{net\_x}\), is the sum of the forces in the X direction: \(F_{net\_x} = F_1x + F_2x = 8~N\) The net force in the Y direction, \(F_{net\_y}\), is the sum of the forces in the Y direction: \(F_{net\_y} = F_1y + F_2y = 6~N\)

Step 2: Find the total net force

We can find the total net force, \(F_{net}\), acting on the object using the Pythagorean theorem for the X and Y components: \(F_{net} = \sqrt{F_{net\_x}^2 + F_{net\_y}^2} = \sqrt{(8~N)^2 + (6~N)^2} = 10~N\)

Step 3: Use Newton's second law to find the acceleration

Newton's second law states that the net force acting on an object is equal to the product of the object's mass and its acceleration: \(F_{net} = m \times a\) We can rearrange the equation and solve for the acceleration: \(a = \frac{F_{net}}{m}\) Plugging in the given values: \(a = \frac{10~N}{5~kg} = 2~\frac{m}{s^2}\) The magnitude of the acceleration of the object is \(2~\frac{m}{s^{2}}\), which corresponds to option (B).

Key Concepts

Net Force Calculation

In order to determine the motion of an object, it is crucial to calculate the net force acting on it. Newton's Second Law provides a relationship that helps link these forces to the object's motion. According to this law, the net force acting on an object is the sum of all individual forces acting on it. Each force is represented as a vector, meaning it has both magnitude and direction.

For this exercise, two forces are given:

• 8 N acting in the X-direction
• 6 N acting in the Y-direction

The net force acting on the object in both the X and the Y directions can be determined by adding the respective components in these directions. Since these forces are at right angles to each other, this process is straightforward:

• Net force in X: 8 N
  
• Net force in Y: 6 N
  

By finding these component forces, we can better understand how the object will move.

Vector Components

The concept of vectors and their components is fundamental in understanding complex problems with force and motion. A vector is a quantity that has both magnitude and direction. In our problem, forces acting in two different directions are vector components.

Each of the given forces can be represented as a vector. Here, it is essential to break down these vectors into their X and Y components.

• For the 8 N force, the X component is 8 N, and the Y component is 0 N
  
• For the 6 N force, the X component is 0 N, and the Y component is 6 N
  

These components tell us how strong the forces are in each direction. By understanding the vector components, we can easily predict how they contribute to the overall net force and motion.

Pythagorean Theorem

The Pythagorean Theorem helps us find the resultant or total net force when forces are acting at right angles to each other. It establishes that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of its other two sides.

In our scenario, to find the total net force on the object, we apply the Pythagorean Theorem to the net forces acting in the X and Y directions. The formula used is:

\[F_{net} = \sqrt{(F_{net\_x})^2 + (F_{net\_y})^2}\]

By substituting the values from the problem:

• \(F_{net\_x} = 8\,N\)
• \(F_{net\_y} = 6\,N\)

We find:
\[F_{net} = \sqrt{(8\,N)^2 + (6\,N)^2} = 10\,N\]

This total net force is then divided by the object's mass to find its acceleration, showcasing the importance of the Pythagorean Theorem in solving such physics problems.