Question

A body of mass 0.5 kg travels on straight line path with velocity v = (3x² + 4) m/s. The network done by the force during its displacement from x = 0 to x = 2 m is

• A. 64 J
• B. 60 J
• C. 120 J
• D. 128 J

Answer 1

The Correct Option is B

The problem involves calculating the work done on a body as it moves over a certain distance. The work done by a force, moving an object from position x_1 to x_2, is given by the work-energy theorem. This can be calculated using the change in kinetic energy formula:

W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2)

Here, v_f and v_i are the final and initial velocities respectively, and m is the mass of the body.

1. First, calculate the initial velocity v_i at x = 0:

Given the velocity function v(x) = 3x^2 + 4, at x = 0:

v_i = 3(0)^2 + 4 = 4 \, \text{m/s}

2. Next, calculate the final velocity v_f at x = 2:

v_f = 3(2)^2 + 4 = 3 \times 4 + 4 = 12 + 4 = 16 \, \text{m/s}

3. Now calculate the change in kinetic energy:

Using the given mass m = 0.5 \, \text{kg}:

\Delta KE = \frac{1}{2} \times 0.5 \times (16^2 - 4^2)

= \frac{1}{2} \times 0.5 \times (256 - 16)

= \frac{1}{2} \times 0.5 \times 240

= \frac{1}{2} \times 120

= 60 \, \text{J}

Thus, the work done by the force during the displacement from x = 0 to x = 2 is 60 J.