Question

A block of mass 100 gm is placed on a smooth surface, moves with the acceleration of a=2x, the change in kinetic energy can be given as \((\frac {x^n}{10})\), find the value of n.

Answer 1

Correct Answer: 2

To solve the problem, we start by noting that the block's mass \(m = 100\, \text{gm} = 0.1\, \text{kg}\) (conversion from grams to kilograms). The acceleration of the block is given as a function of displacement: \(a = 2x\).

Kinetic energy (KE) can be expressed as:

\( \text{KE} = \frac {1}{2}mv^2 \)

Using Newton's second law, \( F = ma \), and integrating to find velocity, we have:

\( F = m \cdot 2x = 0.1 \cdot 2x = 0.2x \)

Work done (WD) by force equals the change in kinetic energy, and can be expressed as

\(\Delta \text{KE} = \int F \, dx = \int 0.2x \, dx \)

Integrating:

\(\Delta \text{KE} = 0.2 \int x \, dx = 0.2 \left( \frac {x^2}{2} \right) = 0.1x^2 \)

We compare this result with the given expression for change in kinetic energy:

\(\Delta \text{KE} = \frac {x^n}{10} \)

Equating the two expressions:

\(\frac {x^n}{10} = 0.1x^2 \)

By comparing powers of \(x\), we find \(n = 2\).

Confirming \(n\) falls within the given range [2,2], we observe that it satisfies this condition exactly. Therefore, the solution is consistent and verified.