Question

A body of mass \(2\) kg is moving with a velocity of \(10\,\text{m s}^{-1}\). If a force of \(50\) N is applied on it for \(10\) s along its motion, the velocity of the body (in \(\text{m s}^{-1}\)) is

• A. \(220\)
• B. \(200\)
• C. \(150\)
• D. \(175\)
• E. \(260\)

Hint:

Always use \(F=ma\) to find acceleration, then apply kinematic equations like \(v=u+at\) when time is given.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem involves the application of Newton's second law of motion and the first equation of kinematics. A constant force is applied to a moving body, which results in a constant acceleration. We can use this acceleration to find the final velocity after a certain time.
Step 2: Key Formula or Approach:
1. Newton's Second Law: The acceleration (\(a\)) produced in a body is directly proportional to the net force (\(F\)) acting on it and inversely proportional to its mass (\(m\)).
\[ F = ma \implies a = \frac{F}{m} \] 2. First Equation of Motion: For an object moving with constant acceleration, the final velocity (\(v\)) is related to the initial velocity (\(u\)), acceleration (\(a\)), and time (\(t\)) by:
\[ v = u + at \] Step 3: Detailed Explanation:
We are given the following values:
- Mass of the body, \( m = 2 \) kg.
- Initial velocity, \( u = 10 \) ms\(^{-1}\).
- Applied force, \( F = 50 \) N.
- Time duration, \( t = 10 \) s.
First, we calculate the acceleration of the body using Newton's second law:
\[ a = \frac{F}{m} = \frac{50 \text{ N}}{2 \text{ kg}} = 25 \text{ ms}^{-2} \] Since the force is applied "along its motion," the acceleration is in the same direction as the initial velocity.
Next, we use the first equation of motion to find the final velocity (\(v\)):
\[ v = u + at \] Substitute the known values:
\[ v = 10 \text{ ms}^{-1} + (25 \text{ ms}^{-2} \times 10 \text{ s}) \] \[ v = 10 \text{ ms}^{-1} + 250 \text{ ms}^{-1} \] \[ v = 260 \text{ ms}^{-1} \] Step 4: Final Answer:
The velocity of the body after 10 seconds is 260 ms\(^{-1}\). This corresponds to option (E).