The displacement x of a body of mass 1 kg on a horizontal smooth surface as a function of time t is given by `x = (t^4)/4`. The work done in the first second is

Options

A. `1/4 J`
B. `1/2 J`
C. `3/4 J`
D. `5/4 J`

Answer

To solve the problem, we need to find the work done on a body of mass 1 kg whose displacement as a function of time is given by \( x = \frac{t^4}{4} \) during the first second (from \( t = 0 \) to \( t = 1 \)).

### Step-by-Step Solution:

1. **Find the Velocity**:
The velocity \( v \) is the first derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^4}{4} \right) = t^3 \]

2. **Find the Acceleration**:
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt} (t^3) = 3t^2 \]

3. **Calculate the Force**:
Using Newton's second law, the force \( F \) acting on the body is given by:
\[ F = m \cdot a \]
where \( m = 1 \, \text{kg} \). Thus,
\[ F = 1 \cdot (3t^2) = 3t^2 \]

4. **Calculate the Work Done**:
The work done \( W \) is given by the integral of force with respect to displacement. We can express this as:
\[ W = \int_{0}^{1} F \, dx \]
To express \( dx \) in terms of \( dt \), we note that:
\[ dx = v \, dt = t^3 \, dt \]
Therefore, substituting for \( F \) and \( dx \):
\[ W = \int_{0}^{1} (3t^2) (t^3) \, dt = \int_{0}^{1} 3t^5 \, dt \]

5. **Evaluate the Integral**:
Now we compute the integral:
\[ W = 3 \int_{0}^{1} t^5 \, dt = 3 \left[ \frac{t^6}{6} \right]_{0}^{1} = 3 \left( \frac{1^6}{6} - \frac{0^6}{6} \right) = 3 \cdot \frac{1}{6} = \frac{1}{2} \, \text{J} \]

Thus, the work done in the first second is \( \frac{1}{2} \, \text{J} \).