© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the given information
• The force acting on the particle is time-dependent and given by $F = 6t$.
• The mass of the particle is $m = 1\,\text{kg}$.
• The particle starts from rest, which means its initial velocity is $v_0 = 0\,\text{m/s}$.
• We need to find the work done by this force in the first 1 second.
Step 2: Relate force to acceleration
Using Newton’s second law, $F = ma$, and noting that $a = \frac{dv}{dt}$, we get:
$ m \frac{dv}{dt} = 6t $
Since $m=1\,\text{kg}$:
$ \frac{dv}{dt} = 6t $
Step 3: Integrate to find velocity
Separate variables and integrate with respect to time $t$ from $0$ to $1\,\text{s}$. Let $v$ be the velocity at time $t$:
$ \int_{0}^{v} dv = \int_{0}^{1} 6t \,dt $
The left-hand side integrates to $v$. The right-hand side integrates to:
$ 6 \int_{0}^{1} t \,dt = 6 \left[ \frac{t^2}{2} \right]_{0}^{1} = 6 \cdot \frac{1^2}{2} = 3 $
Thus, at $t = 1\,\text{s}$:
$ v = 3\,\text{m/s} $
Step 4: Use the work-energy theorem
The work done by the force equals the change in kinetic energy of the particle. Since the particle starts from rest, its initial kinetic energy is zero. After 1 second, its kinetic energy is:
$ \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \cdot 1 \cdot (3)^2 = \frac{1}{2} \cdot 9 = 4.5\,\text{J} $
Step 5: State the final answer
Therefore, the work done by the force during the first 1 second is $4.5\,\text{J}$.
