A body of mass `0.5 kg` travels in a straight line with velocity `v =a x^(3//2)` where `a = 5 m//s^(2)`. The work done by the net force during its displacement from `x = 0` to `x = 2 m` is

Options

A. 1.5 J
B. 50 J
C. 10 J
D. 100 J

Explanation

To solve the problem of calculating the work done by the net force on a body of mass \(0.5 \, \text{kg}\) traveling with a velocity given by \(v = a x^{3/2}\) (where \(a = 5 \, \text{m/s}^2\)) during its displacement from \(x = 0\) to \(x = 2 \, \text{m}\), we can follow these steps:

### Step 1: Determine the expression for velocity
The velocity of the body is given as:
\[ v = a x^{3/2} \]
Substituting the value of \(a\):
\[ v = 5 x^{3/2} \]

### Step 2: Find the acceleration
Acceleration \(a_0\) can be expressed as:
\[ a_0 = \frac{dv}{dt} \]
Using the chain rule:
\[ a_0 = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Now, we need to differentiate \(v\) with respect to \(x\):
\[ \frac{dv}{dx} = \frac{d}{dx}(5 x^{3/2}) = 5 \cdot \frac{3}{2} x^{1/2} = \frac{15}{2} x^{1/2} \]
Thus, substituting back:
\[ a_0 = v \cdot \frac{dv}{dx} = (5 x^{3/2}) \cdot \left(\frac{15}{2} x^{1/2}\right) = \frac{75}{2} x^2 \]

### Step 3: Calculate the force
Using Newton's second law, the force \(F\) can be calculated as:
\[ F = m a_0 \]
Substituting the mass \(m = 0.5 \, \text{kg}\):
\[ F = 0.5 \cdot \frac{75}{2} x^2 = \frac{75}{4} x^2 \]

### Step 4: Calculate the work done
The work done \(W\) by the net force during the displacement from \(x = 0\) to \(x = 2\) is given by the integral:
\[ W = \int_{0}^{2} F \, dx = \int_{0}^{2} \left(\frac{75}{4} x^2\right) dx \]
Calculating the integral:
\[ W = \frac{75}{4} \int_{0}^{2} x^2 \, dx = \frac{75}{4} \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{75}{4} \cdot \frac{2^3}{3} = \frac{75}{4} \cdot \frac{8}{3} = \frac{600}{12} = 50 \, \text{J} \]

### Final Answer
The work done by the net force during the displacement from \(x = 0\) to \(x = 2 \, \text{m}\) is:
\[ \boxed{50 \, \text{J}} \]