Question

A body of mass 0.5 kg travels in a straight line with velocity V = ax3/2 where a = 5 m1/2s1. Find the work done by the net force during its displacement from x = 0 to x = 2 m is :-
(1)1.5 J
(2)50 J
(3)10 J
(4)100 J

Answer 1

Answer on Question#40690 – Physics – Mechanics

A body of mass $0.5\,\mathrm{kg}$ travels in a straight line with velocity $V = ax^3 / 2$ where $a = 5\,\mathrm{m}^1/\mathrm{s}^{-1}$. Find the work done by the net force during its displacement from $x = 0$ to $x = 2\,\mathrm{m}$ is :-

(1) $1.5\,\mathrm{J}$

(2) $50\,\mathrm{J}$

(3) $10\,\mathrm{J}$

(4) $100\,\mathrm{J}$

Solution:

$\mathrm{m} = 0.5\,\mathrm{kg}$ – mass of the body;

Here the unit of the constant $a$ is not correct, a should have dimension $\mathrm{L}^{\frac{1}{2}}\,\mathrm{T}^{-1}$ and not $\mathrm{L}\,\mathrm{T}^{-2}$.

Velocity is given by $\left(a = 5\,\frac{\mathrm{m}}{\mathrm{s}^2}\right)$:

$$v = \frac{dx}{dt} = a x^{\frac{3}{2}}$$

Let $a' =$ acceleration:

$$a' = \frac{dv}{dt} \Rightarrow a' = \frac{d\left(ax^{\frac{3}{2}}\right)}{dt} = \frac{3a}{2} x^{\frac{1}{2}} \frac{dx}{dt}$$ $$a' = \frac{3a}{2} x^{\frac{1}{2}} a x^{\frac{3}{2}} = \frac{3a^2}{2} x^3$$

Newton's second Law for the body:

$$F = m a'$$ $$(3)\text{to}(4):$$ $$F = m \frac{3a^2}{2} x^3$$

The work done by the net force during its displacement:

$$W = \int_{x = x_1}^{x = x_2} F \, dx$$ $$W = \int_{x = x_1}^{x = x_2} m \frac{3a^2}{2} x^3 \, dx = m \frac{3a^2}{2} \cdot \frac{x^4}{4} \int_{x=0}^{x=2} a^2 \, dx = \frac{3a^2 m}{2} \left(\frac{2^4}{4} - 0\right) = 6a^2 m$$ $$= 6 \cdot \left(5\,\frac{\mathrm{m}}{\mathrm{s}^2}\right)^2 \cdot 0.5\,\mathrm{kg} = 75\,\mathrm{J}$$

Answer: work done by the net force is equal to 75 J

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