A body of mass $2\, kg$ is initially at rest It starts moving unidirectionally under the influence of a source of constant power P Its displacement in $4 s$ is $\frac{1}{3} \alpha^2 \sqrt{P} m$ The value of $\alpha$ will be_______

Question

A body of mass $4$ kg is placed at a point $P$ having coordinates $ (3,4) $ m. Under the action of force $ \mathbf{F} = (2\hat{i} + 3\hat{j}) $ N, it moves to a new point $Q$ having coordinates $ (6,10) $ m in $4$ sec. The average power and instantaneous power at the end of $4$ sec are in the ratio:

Answer 1

To solve the problem, we first relate power, kinetic energy, and displacement. The power $ P $ imparted to the body results in kinetic energy $ E $ given by:

$ E = \frac{1}{2}mv^2 $

Since the body starts from rest, the work done in time $ t $ is equal to the kinetic energy, and for constant power:

$ P \times t = \frac{1}{2}mv^2 $

Rearranging for velocity $ v $, we have:

$ v = \sqrt{\frac{2Pt}{m}} $

The displacement $ s $ under constant acceleration is given by:

$ s = \int v \, dt = \int \sqrt{\frac{2Pt}{m}} \, dt $

To find this, evaluate:

$ s = \frac{1}{3} \sqrt{\frac{2P}{m}} \, t^{3/2} \bigg|_{0}^{4} $

Substituting values $ m = 2 \, kg $, $ t = 4 \, s $, and equating to the given expression for displacement:

$ \frac{1}{3} \alpha^2 \sqrt{P} = \frac{1}{3} \sqrt{\frac{2P}{2}} \times 4^{3/2} $

Simplify:

$ \alpha^2 \sqrt{P} = \sqrt{P} \times 8 $

Equating coefficients of $\sqrt{P}$:

$ \alpha^2 = 8 $

Solving for $\alpha$:

$ \alpha = \sqrt{8} = 2\sqrt{2} \approx 4 $

Verify that $\alpha$ ($\approx 4$) is within the given range (4, 4). Conclude the solution, $\alpha = 4$, is valid.

A body of mass $2\, kg$ is initially at rest It starts moving unidirectionally under the influence of a source of constant power P Its displacement in $4 s$ is $\frac{1}{3} \alpha^2 \sqrt{P} m$ The value of $\alpha$ will be_______

Correct Answer: 4

Solution and Explanation

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