Under the action of a force, a 1 kg body moves, such that its position x as function of time t is given by `x=(t^(3))/(2).` where x is in meter and t is in second. The work done by the force in fiest 3 second is

Options

A. `143 J`
B. `2430 J`
C. `(729)/(8)J`
D. `24.3 J`

Answer

To find the work done by the force on a 1 kg body moving according to the position function \( x(t) = \frac{t^3}{2} \), we will follow these steps:

### Step 1: Find the velocity
The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{2}\right) = \frac{3t^2}{2} \]

### Step 2: Find the acceleration
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \) with respect to time \( t \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{3t^2}{2}\right) = 3t \]

### Step 3: Find the force
Using Newton's second law, the force \( F \) acting on the body can be calculated as:

\[ F = m \cdot a = 1 \, \text{kg} \cdot 3t = 3t \, \text{N} \]

### Step 4: Calculate the work done
The work done \( W \) by the force over the time interval from \( t = 0 \) to \( t = 3 \) seconds can be expressed as:

\[ W = \int_{0}^{3} F \cdot v(t) \, dt \]

Substituting \( F \) and \( v(t) \):

\[ W = \int_{0}^{3} (3t) \left(\frac{3t^2}{2}\right) dt \]

This simplifies to:

\[ W = \int_{0}^{3} \frac{9t^3}{2} \, dt \]

### Step 5: Evaluate the integral
Now we evaluate the integral:

\[ W = \frac{9}{2} \int_{0}^{3} t^3 \, dt = \frac{9}{2} \left[\frac{t^4}{4}\right]_{0}^{3} \]

Calculating the definite integral:

\[ W = \frac{9}{2} \cdot \left(\frac{3^4}{4} - \frac{0^4}{4}\right) = \frac{9}{2} \cdot \left(\frac{81}{4}\right) \]

\[ W = \frac{9 \cdot 81}{8} = \frac{729}{8} \, \text{J} \]

### Final Answer
Thus, the work done by the force in the first 3 seconds is:

\[ W = 91.125 \, \text{J} \]