Calculate Work Done on Body with Position Dependent Velocity

To find the work done on the body during its displacement from $x = 0$ to $x = 5\text{ m}$, we can apply the Work-Energy Theorem, which states that the net work done on an object is equal to the change in its kinetic energy:

$W = \Delta K = K_f - K_i = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2$

Given:

• Mass of the body, $m = 1\text{ kg}$
• Velocity as a function of position, $v(x) = 2x^2$

Step 1: Calculate the initial velocity ($v_i$) at $x = 0\text{ m}$ $v_i = v(0) = 2(0)^2 = 0\text{ m/s}$

Step 2: Calculate the final velocity ($v_f$) at $x = 5\text{ m}$ $v_f = v(5) = 2(5)^2 = 2 \times 25 = 50\text{ m/s}$

Step 3: Calculate the work done ($W$) $W = \frac{1}{2} m \left(v_f^2 - v_i^2\right)$ $W = \frac{1}{2} \times 1\text{ kg} \times \left(50^2 - 0^2\right)$ $W = \frac{1}{2} \times 2500 = 1250\text{ J}$

(Alternatively, using integration of force:) $a = v \frac{dv}{dx} = (2x^2) \frac{d}{dx}(2x^2) = (2x^2)(4x) = 8x^3\text{ m/s}^2$ $F = m a = 1 \times 8x^3 = 8x^3\text{ N}$ $W = \int_{0}^{5} F \, dx = \int_{0}^{5} 8x^3 \, dx = \left[ 2x^4 \right]_{0}^{5} = 2(5)^4 - 0 = 2 \times 625 = 1250\text{ J}$

Thus, the work done by the force acting on the body is $1250\text{ J}$.

Correct Answer: C