Problem 65 - A Complete Resource Book in Physics for JEE Main

Question

A constant force acting on a body of mass \(3.0 \mathrm{~kg}\) changes its speed from \(2.0 \mathrm{~ms}^{-1}\) to \(3.5 \mathrm{~ms}^{-1}\) in \(25 \mathrm{~s}\). The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? (A) \(0.18 \mathrm{~N}\) (B) \(0.36 \mathrm{~N}\) (C) \(0.9 \mathrm{~N}\) (D) None of these

Step-by-Step Solution

Answer

The magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. The correct answer is (A).

Step 1: Find acceleration

We need to find acceleration (a) first, which can be calculated by subtracting the initial velocity (u) from the final velocity (v) and dividing the result by the time (t) taken for the change: \[a = \frac{v - u}{t}\] Using the given values in the problem, we have: \[a = \frac{3.5 \mathrm{~ms}^{-1} - 2.0 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\]

Step 2: Calculate acceleration

Now, we calculate the acceleration: \[a = \frac{1.5 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\] \[a = 0.06 \mathrm{~ms}^{-2}\]

Step 3: Apply Newton's second law

Now, we can use Newton's second law of motion to find the force (F) acting on the object, using the mass (m) and the acceleration (a): \[F = m \cdot a\] Plugging in the given mass and the calculated acceleration: \[F = 3.0 \mathrm{~kg} \cdot 0.06 \mathrm{~ms}^{-2}\]

Step 4: Calculate the force

Finally, we calculate the force: \[F = 0.18 \mathrm{~N}\]

Step 5: Determine the direction of the force

Since the direction of motion of the body remains unchanged, the force must be acting in the same direction as the initial velocity. Therefore, the magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. So the correct answer is (A).