A particle of mass 2 kg starts motion at time t = 0 under the action of variable force F = 4t (where F is in N and t is in s). The work done by this force in first two second is

Options

A. 16 J
B. 8 J
C. 4 J
D. Zero

Answer

To solve the problem of finding the work done by a variable force on a particle of mass 2 kg over the first two seconds, we can follow these steps:

### Step 1: Understand the Force
The force acting on the particle is given by the equation:
\[ F(t) = 4t \]
where \( F \) is in Newtons and \( t \) is in seconds.

### Step 2: Calculate the Acceleration
Using Newton's second law, the acceleration \( a \) can be calculated as:
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the particle. Substituting the values:
\[ a = \frac{4t}{2} = 2t \]

### Step 3: Relate Acceleration to Velocity
We know that acceleration is the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = 2t \]

### Step 4: Integrate to Find Velocity
To find the velocity \( v \), we integrate both sides with respect to \( t \):
\[ \int dv = \int 2t \, dt \]
This gives us:
\[ v = t^2 + C \]
where \( C \) is the constant of integration. Since the particle starts from rest at \( t = 0 \), we have:
\[ v(0) = 0 \Rightarrow C = 0 \]
Thus, the velocity function is:
\[ v(t) = t^2 \]

### Step 5: Calculate the Velocity at \( t = 2 \) seconds
Now, we can find the velocity at \( t = 2 \) seconds:
\[ v(2) = (2)^2 = 4 \, \text{m/s} \]

### Step 6: Calculate the Change in Kinetic Energy
The work done by the force is equal to the change in kinetic energy. The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2}mv^2 \]
Calculating the initial kinetic energy at \( t = 0 \):
\[ KE_{initial} = \frac{1}{2} \cdot 2 \cdot 0^2 = 0 \]
Calculating the kinetic energy at \( t = 2 \):
\[ KE_{final} = \frac{1}{2} \cdot 2 \cdot (4)^2 = \frac{1}{2} \cdot 2 \cdot 16 = 16 \, \text{J} \]

### Step 7: Calculate the Work Done
The work done \( W \) is the change in kinetic energy:
\[ W = KE_{final} - KE_{initial} = 16 - 0 = 16 \, \text{J} \]

### Final Answer
The work done by the force in the first two seconds is:
\[ \boxed{16 \, \text{J}} \]

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