Ncert Exercises & Solutions

A body of mass $0.5$ kg travels in a straight line with velocity $v=ax^{3/2}$ where $a=5~\mathrm{m^{-1/2}s^{-1}}$. The work done by the net force during its displacement from $x=0$ m to $x=2$ m is:
1. $15$ J
2. $50$ J
3. $10$ J
4. $100$ J

(b)

Hint: Using the equation of the velocity, we can find the acceleration.

Step 1: Find the acceleration.

$\begin{array}{l} Given, v = {ax}^{3 / 2} \\ m = 0 .5 kg, a = 5 m^{- 1 / 2} s^{- 1}, work done (W) = ? \end{array}$

We know that:

Acceleration

$\begin{array}{l} a_{0} = \frac{dv}{dt} = v \frac{dv}{dx} = {ax}^{3 / 2} \frac{d}{dx} ({ax}^{3 / 2})^{'} \\ = {ax}^{3 / 2} \times a \times \frac{3}{2} \times x^{1 / 2} = \frac{3}{2} a^{2} x^{2} \end{array}$

Step 2: Find the force.

Now, $Force = {ma}_{0} = m \frac{3}{2} a^{2} x^{2}$

Step 2: Find the work done.

$\begin{array}{l} Work done = \int_{x = 0}^{x = 2} Fdx = \int_{0}^{2} \frac{3 {ma}^{2} x^{2} dx}{2} \\ = \frac{3}{2} {ma}^{2} \times (x^{3} / 3)_{0}^{2} \\ = \frac{1}{2} {ma}^{2} \times 8 = \frac{1}{2} \times (0 .5) \times (25) \times 8 = 50 J \end{array}$

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