Question

A body of mass \(1\,\text{kg}\) moves along a straight line with a velocity \(v=2x^2\). The work done by the body during displacement from \(x=0\) to \(x=5\,\text{m}\) is _____ J.

• A. \(0\)
• B. \(250\)
• C. \(1250\)
• D. \(1000\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question
We are given the velocity of a body as a function of its position. We need to calculate the work done on the body as it moves from one position to another. The work-energy theorem is the most direct way to solve this.
Step 2: Key Formula or Approach
The Work-Energy Theorem states that the work done (\(W\)) on an object by the net force is equal to the change in its kinetic energy (\(\Delta K\)). \[ W = \Delta K = K_{final} - K_{initial} \] where the kinetic energy \(K\) is given by \( K = \frac{1}{2}mv^2 \).
Step 3: Detailed Explanation
We are given: - Mass of the body, \( m = 1 \) kg. - Velocity as a function of position, \( v(x) = 2x^2 \). - Initial position, \( x_{initial} = 0 \) m. - Final position, \( x_{final} = 5 \) m. First, we calculate the initial and final velocities. Initial velocity at \( x = 0 \): \[ v_{initial} = v(0) = 2(0)^2 = 0 \text{ m/s} \] Final velocity at \( x = 5 \): \[ v_{final} = v(5) = 2(5)^2 = 2(25) = 50 \text{ m/s} \] Next, we calculate the initial and final kinetic energies. Initial kinetic energy: \[ K_{initial} = \frac{1}{2}mv_{initial}^2 = \frac{1}{2}(1)(0)^2 = 0 \text{ J} \] Final kinetic energy: \[ K_{final} = \frac{1}{2}mv_{final}^2 = \frac{1}{2}(1)(50)^2 = \frac{1}{2}(2500) = 1250 \text{ J} \] Finally, we apply the work-energy theorem to find the work done. \[ W = K_{final} - K_{initial} = 1250 \text{ J} - 0 \text{ J} = 1250 \text{ J} \] Alternative Method (using integration of force): We can find the force by first finding the acceleration. \[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \] Given \( v = 2x^2 \), we have \( \frac{dv}{dx} = 4x \). \[ a(x) = (2x^2)(4x) = 8x^3 \] The force is \( F(x) = ma(x) = (1)(8x^3) = 8x^3 \). Work done is the integral of force over displacement: \[ W = \int_{x_{initial}}^{x_{final}} F(x) dx = \int_{0}^{5} 8x^3 dx \] \[ W = \left[ \frac{8x^4}{4} \right]_0^5 = \left[ 2x^4 \right]_0^5 = 2(5)^4 - 2(0)^4 = 2(625) = 1250 \text{ J} \] Both methods yield the same result. Step 4: Final Answer
The work done by the body is 1250 J.