A body of mass 0.5 kg travels in a straight line with velocity v =a x $^{3/2}$ where a = 5 m$^{-1/2}$ s–1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Question

A wire of resistance $ 9 \, \Omega $ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be:

Answer 1

Given Data

Mass: $m = 0.5$ kg
Velocity: $v = a x^{3/2}$
Constant: $a = 5$ m$^{-1/2}$ s$^{-1}$
Displacement: $x = 0$ to $x = 2$ m

Work-Energy Theorem

Net work = change in kinetic energy:

$$W_\text{net} = \Delta K = K_f - K_i = \frac{1}{2} m (v_f^2 - v_i^2)$$

Velocity Calculation

$$v_i = a (0)^{3/2} = 0 \, \text{m/s}$$
$$v_f = 5 \times (2)^{3/2} = 5 \times (2\sqrt{2}) = 5 \times 2.828 = 14.14 \, \text{m/s}$$

$$v_f^2 = (14.14)^2 = 200 \, \text{m}^2\text{/s}^2$$

Work Done

$$W_\text{net} = \frac{1}{2} \times 0.5 \times (200 - 0) = 0.25 \times 200 = 50 \, \text{J}$$

Work Done by Net Force

$W = \textbf{50 J}$

Verification (Force Method)

Acceleration: $a_x = \frac{dv}{dt} = v \frac{dv}{dx}$

$$v = 5 x^{3/2}, \quad \frac{dv}{dx} = \frac{15}{2} x^{1/2}$$ $$F = m v \frac{dv}{dx} = 0.5 \times 5 x^{3/2} \times \frac{15}{2} x^{1/2} = \frac{75}{4} x^2$$

$$W = \int_0^2 F \, dx = \frac{75}{4} \int_0^2 x^2 \, dx = \frac{75}{4} \times \frac{8}{3} = 50 \, \text{J} \quad \checkmark$$

Key Insight

Work-energy theorem simplifies variable force problems. Direct $\Delta KE$ always works!

A body of mass 0.5 kg travels in a straight line with velocity v =a x \(^{3/2}\) where a = 5 m\(^{-1/2}\) s–1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Solution and Explanation