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

Options

A. `vec(E) oo -B_(0)c"sin" (kx +omegat)hat(k)V//m`
B. `vec(E) = B_(0)c"sin"(kx - omegat)hat(k)V//m`
C. `vec(E) = B_(0)c"sin"(kx + omegat)hat(k) V//m`
D. `vec(E) = (B_(0))/(c)"sin" (kx + omegat) hat(k)V//m`

Answer

To solve the problem of finding the work done by the force \( F = 6t \) on a particle of mass \( 1 \, \text{kg} \) during the first second, we can follow these steps:

### Step 1: Identify the force and mass
Given:
- Force \( F(t) = 6t \)
- Mass \( m = 1 \, \text{kg} \)

### Step 2: Calculate acceleration
Using Newton's second law, we know that:
\[ F = m \cdot a \]
Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} = \frac{6t}{1} = 6t \]

### Step 3: Relate acceleration to velocity
Acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
So we can write:
\[ \frac{dv}{dt} = 6t \]

### Step 4: Integrate to find velocity
Integrate both sides with respect to time from \( t = 0 \) to \( t = 1 \):
\[ \int_{0}^{v} dv = \int_{0}^{1} 6t \, dt \]
The left side integrates to \( v \), and the right side evaluates as follows:
\[ \int_{0}^{1} 6t \, dt = 6 \cdot \frac{t^2}{2} \Big|_{0}^{1} = 6 \cdot \frac{1^2}{2} - 6 \cdot \frac{0^2}{2} = 3 \]
Thus, we find:
\[ v = 3 \, \text{m/s} \]

### Step 5: Calculate work done
The work done \( W \) is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Since the particle starts from rest, the initial kinetic energy is zero:
\[ KE_{\text{initial}} = 0 \]
The final kinetic energy is given by:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 1 \cdot (3)^2 = \frac{1}{2} \cdot 1 \cdot 9 = \frac{9}{2} \, \text{J} \]
Thus, the work done by the force during the first second is:
\[ W = \frac{9}{2} \, \text{J} \]

### Final Answer
The work done by the force during the first second is \( \frac{9}{2} \, \text{J} \).

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