Ncert Exercises & Solutions

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

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

Step 1: Find the acceleration.

\begin{matrix} \end{matrix}

$\begin{array}{l} \end{array}$

\begin{matrix} \end{matrix}

$\begin{array}{l} \end{array}$

Given,

$v = \left(ax\right)^{3 / 2} \\ m = 0 .5~\text{kg} , a = 5 ~\text{m}^{- 1 / 2} \text{s}^{- 1}$

We know that the acceleration is given by

$a_{0} = \frac{dv}{dt} = v \frac{dv}{dx} = \left(ax\right)^{3 / 2} \frac{d}{dx} \left(ax\right)^{3 / 2} \\ = \left(ax\right)^{3 / 2} \times a \times \frac{3}{2} \times x^{1 / 2} = \frac{3}{2} a^{2} x^{2}$

$\begin{array}{l} \end{array}$

Step 2: Find the work done.

$\text{Work done}= \int_{x = 0}^{x = 2} Fdx = \int_{0}^{2} \frac{3 \left(ma^{2} \right)x^{2} dx}{2} \\ = \frac{3}{2} \left(ma^{2} \right)\times \left[\right. x^{3} / 3 \left.\right]_{0}^{2} \\ = \frac{1}{2} \left(ma^{2}\right) \times 8 = \frac{1}{2} \times \left(\right. 0 .5 \left.\right) \times \left(\right. 25 \left.\right) \times 8 = 50 ~\text J$

Hence, option (2) is the correct answer.

$\begin{array}{l} \end{array}$

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