Question

A time dependent force $F = 6t$ acts on a particle of mass $1\,kg$. If the particle starts from rest, the work done by the force during the first $1\,sec$. will be :

• A. 4.5 J
• B. 22 J
• C. 9 J
• D. 18 J

Answer 1

The Correct Option is A

To determine the work done by the force on the particle during the first second, we first need to understand the relationship between force, motion, and work done.

1. Calculate the Force: The force acting on the particle is given as a time-dependent function \( F = 6t \).
2. Determine Velocity as a Function of Time: Since the particle starts from rest, its initial velocity \( v_0 = 0 \). Using Newton's second law, \( F = ma \), and knowing that \( F = 6t \) and \( m = 1 \, \text{kg} \), the acceleration \( a = F/m = 6t/1 = 6t \).
   • We find velocity by integrating acceleration: \(\frac{dv}{dt} = 6t\).
   • \(\int dv = \int 6t \, dt \Rightarrow v = 3t^2 + C\).
   • Given \( v_0 = 0 \) at \( t = 0 \), we find \( C = 0 \). Thus, \( v = 3t^2 \).
3. Calculate the Work Done: The work done by a force is given by the integral of \( F \cdot ds \), where \( ds \) is the displacement. Since \( v = \frac{ds}{dt} \), the displacement \( ds = v \, dt = 3t^2 \, dt \).
   • Work done, \( W = \int F \cdot ds = \int_0^1 6t \cdot 3t^2 \, dt = \int_0^1 18t^3 \, dt\).
   • \(W = [\frac{18t^4}{4}]_0^1 = \frac{18}{4} [1^4 - 0^4] = 4.5 \, \text{J}\).

Thus, the work done by the force during the first second is 4.5 J, which matches the correct answer.