Question

A body of mass 0.25 kg travels along a straight line from \( x = 0 \) to \( x = 2 \, \text{m} \) with a speed \( v = k x^2 \) where \( k = 2 \, \text{m}^{-1} \). The work done by the net force during this displacement is

• A. 32 J
• B. 4 J
• C. 1 J
• D. 16 J

Hint:

When calculating the work done by a force, the change in kinetic energy is often the easiest method to apply, especially when velocity is given as a function of position.

Answer 1

The Correct Option is B

The speed \( v \) is given by the equation \( v = k x^2 \), where \( k \) is a constant. To find the work done by the net force, we use the work-energy theorem, which states that the work done by the net force is equal to the change in kinetic energy. The kinetic energy \( K \) of the body at position \( x \) is given by: \[ K = \frac{1}{2} m v^2 \] Substituting the expression for \( v \): \[ K = \frac{1}{2} m (k x^2)^2 = \frac{1}{2} m k^2 x^4 \] Now, the work done by the net force is the change in kinetic energy as the body moves from \( x = 0 \) to \( x = 2 \, \text{m} \). The change in kinetic energy is given by: \[ W = K_2 - K_1 = \frac{1}{2} m k^2 x_2^4 - 0 \] Substitute \( m = 0.25 \, \text{kg} \), \( k = 2 \, \text{m}^{-1} \), and \( x_2 = 2 \, \text{m} \): \[ W = \frac{1}{2} (0.25) (2^2)^4 = \frac{1}{2} (0.25) \times 16 = 4 \, \text{J} \] Thus, the work done by the net force is \( 4 \, \text{J} \).