Question

A body of mass \( 0.2\,\text{kg} \) travels along a straight line with velocity \( v = (2x^2 + 2)\,\text{m s}^{-1} \). The net work done by the driving force during its displacement from \( x = 0 \) to \( x = 2 \,\text{m} \) is

• A. \(2\,\text{J} \)
• B. \(5.4\,\text{J} \)
• C. \(9.6\,\text{J} \)
• D. \(10.8\,\text{J} \)
• E. \(6.5\,\text{J} \)

Hint:

Always use kinetic energy change for work in variable velocity problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem can be solved using the Work-Energy Theorem. The theorem states that the net work done on an object is equal to the change in its kinetic energy.
Step 2: Key Formula or Approach:
1. Work-Energy Theorem: \(W_{net} = \Delta K = K_f - K_i\), where \(K = \frac{1}{2}mv^2\) is the kinetic energy. 2. Calculate the initial velocity \(v_i\) at \(x=0\). 3. Calculate the final velocity \(v_f\) at \(x=2\). 4. Calculate the initial kinetic energy \(K_i\) and the final kinetic energy \(K_f\). 5. Find the difference to get the net work done.
Step 3: Detailed Explanation:
The mass of the body is \(m = 0.2\) kg. The velocity is given by the function \(v(x) = 2x^2 + 2\). 1. Calculate initial and final velocities. Initial position: \(x_i = 0\) m. Initial velocity: \(v_i = v(0) = 2(0)^2 + 2 = 2\) m/s. Final position: \(x_f = 2\) m. Final velocity: \(v_f = v(2) = 2(2)^2 + 2 = 2(4) + 2 = 8 + 2 = 10\) m/s. 2. Calculate initial and final kinetic energies. Initial kinetic energy: \[ K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(0.2)(2)^2 = \frac{1}{2}(0.2)(4) = 0.4 \text{ J} \] Final kinetic energy: \[ K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(0.2)(10)^2 = \frac{1}{2}(0.2)(100) = 10 \text{ J} \] 3. Calculate the net work done. \[ W_{net} = K_f - K_i = 10 - 0.4 = 9.6 \text{ J} \] Step 4: Final Answer:
The net work done is 9.6 J.