Question

A bead of mass \(400\ \text{g}\) is moving along a straight line under a force that delivers a constant power \(1.2\ \text{W}\) to the bead. If the bead is initially at rest, the speed it attains after \(6\ \text{s}\) in \(\text{m s}^{-1}\) is

• A. \(5\)
• B. \(4\)
• C. \(6\)
• D. \(3\)

Hint:

When constant power is supplied, use \(W=Pt\), and if the object starts from rest, equate it to \(\frac12mv^2\).

Answer 1

The Correct Option is C

Step 1: Convert mass into SI unit.
Given mass is
\[ 400\ \text{g} \]
Since
\[ 1000\ \text{g}=1\ \text{kg} \]
we get
\[ 400\ \text{g}=0.4\ \text{kg} \]

Step 2: Use the relation between power, work and kinetic energy.
Power is defined as work done per unit time.
So,
\[ P=\frac{W}{t} \]
Therefore, work done in time \(t\) is
\[ W=Pt \]
Given,
\[ P=1.2\ \text{W},\qquad t=6\ \text{s} \]
Thus,
\[ W=1.2\times6 \]
\[ W=7.2\ \text{J} \]

Step 3: Apply work-energy theorem.
Since the bead is initially at rest, its initial kinetic energy is zero.
The work done becomes the final kinetic energy.
Hence,
\[ W=\frac12mv^2 \]
Substitute the values:
\[ 7.2=\frac12(0.4)v^2 \]
\[ 7.2=0.2v^2 \]
\[ v^2=\frac{7.2}{0.2} \]
\[ v^2=36 \]
\[ v=6\ \text{m s}^{-1} \]

Step 4: Final conclusion.
Hence, the speed attained by the bead after \(6\ \text{s}\) is
\[ \boxed{6\ \text{m s}^{-1}} \]