Question

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

Answer 1

We are given that a block of mass \(m = 100 \, \text{g} = 0.1 \, \text{kg}\) moves on a smooth surface with an acceleration \(a = 2x\), where \(x\) is the position. We need to determine the change in kinetic energy, expressed as \( \left( \frac{x^n}{10} \right) \), and find the value of \(n\). 

Step 1: Relate acceleration and velocity.
The acceleration is given by \( a = 2x \). We know that acceleration is the rate of change of velocity, so we can write: \[ a = \frac{dv}{dt} \] Since \(a = 2x\), this implies: \[ \frac{dv}{dt} = 2x \] 

Step 2: Use the chain rule to relate velocity and position.
Since \(v = \frac{dx}{dt}\), we can use the chain rule to write the expression for acceleration: \[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \] Therefore, the equation becomes: \[ \frac{dv}{dx} \cdot v = 2x \] 

Step 3: Solve the differential equation.
Rearranging the terms, we get: \[ \frac{dv}{dx} = \frac{2x}{v} \] Now, multiply both sides by \(v\) and integrate: \[ v \, dv = 2x \, dx \] Integrating both sides: \[ \int v \, dv = \int 2x \, dx \] This gives: \[ \frac{v^2}{2} = x^2 + C \] where \(C\) is the constant of integration. We can determine \(C\) by setting initial conditions. If the block starts from rest, \(v = 0\) when \(x = 0\). Substituting these values: \[ \frac{0^2}{2} = 0^2 + C \implies C = 0 \] So the equation becomes: \[ \frac{v^2}{2} = x^2 \] or \[ v^2 = 2x^2 \] 

Step 4: Calculate the change in kinetic energy.
The kinetic energy is given by: \[ KE = \frac{1}{2} m v^2 \] Substituting \(v^2 = 2x^2\) and \(m = 0.1 \, \text{kg}\): \[ KE = \frac{1}{2} \times 0.1 \times 2x^2 = 0.1 x^2 \] The change in kinetic energy is the difference in kinetic energy from \(x = 0\) to a general \(x\). Initially, when \(x = 0\), \(KE = 0\), so the change in kinetic energy is: \[ \Delta KE = 0.1 x^2 \] 

Step 5: Match the given form of the change in kinetic energy.
We are given that the change in kinetic energy is of the form: \[ \Delta KE = \frac{x^n}{10} \] Equating the two expressions: \[ 0.1 x^2 = \frac{x^n}{10} \] Multiplying both sides by 10: \[ x^2 = x^n \] Therefore, \(n = 2\). 

Final Answer:
The value of \(n\) is \( \boxed{2} \).