In one dimensional motin, a 1 kg body experiences a force, which is linear function of time t viz. `F=2t` acting in the direction of motion. What is the work done by the force in the first 4 second?

Options

A. 16J
B. 32J
C. 64J
D. 128J

Answer

To solve the problem of calculating the work done by a force that is a linear function of time, we can follow these steps:

### Step 1: Understand the Force Function
The force acting on the body is given by the equation:
\[ F(t) = 2t \]
This means that the force changes linearly with time.

### Step 2: Relate Force to Acceleration
Using Newton's second law, we know that:
\[ F = ma \]
where \( m \) is the mass of the body (1 kg) and \( a \) is the acceleration. Therefore, we can express acceleration as:
\[ a = \frac{F}{m} = \frac{2t}{1} = 2t \]

### Step 3: Relate Acceleration to Velocity
Acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = 2t \]

### Step 4: Integrate to Find Velocity
To find the velocity as a function of time, we integrate the acceleration:
\[ dv = 2t \, dt \]
Integrating both sides from \( t = 0 \) to \( t = t \):
\[ \int dv = \int 2t \, dt \]
This gives:
\[ v(t) = t^2 + C \]
Since the initial velocity \( v(0) = 0 \), we have \( C = 0 \). Therefore:
\[ v(t) = t^2 \]

### Step 5: Calculate Work Done
The work done by the force over a time interval can be calculated using the integral:
\[ W = \int F \cdot ds \]
Since \( ds = v \, dt \), we can rewrite this as:
\[ W = \int F(t) v(t) \, dt \]
Substituting the expressions for \( F(t) \) and \( v(t) \):
\[ W = \int_0^4 (2t)(t^2) \, dt \]
This simplifies to:
\[ W = \int_0^4 2t^3 \, dt \]

### Step 6: Evaluate the Integral
Now we evaluate the integral:
\[ W = 2 \int_0^4 t^3 \, dt \]
Calculating the integral:
\[ \int t^3 \, dt = \frac{t^4}{4} \]
Thus:
\[ W = 2 \left[ \frac{t^4}{4} \right]_0^4 = 2 \left[ \frac{4^4}{4} - \frac{0^4}{4} \right] \]
Calculating \( 4^4 = 256 \):
\[ W = 2 \left[ \frac{256}{4} \right] = 2 \times 64 = 128 \, \text{Joules} \]

### Final Answer
The work done by the force in the first 4 seconds is:
\[ \boxed{128 \, \text{Joules}} \]