Question

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

• A. 50 J
• B. 10 J
• C. 20 J
• D. 30 J

Answer 1

Answer: (A)

Mass of the body (m) = 0.5 kg
Velocity of the body (v) = ax^3/2
where, a = 5m^–1/2/s
Velocity of the body at x = 0,
v₁ = 5 $\times$ 0 = 0
Velocity of the body at c = 2 m
v₂ = 5 $\times$ (2)^3/2m/s
According to work-energy theorem
Work done = Change in K.E.
$= \frac{1}{2} \text{mv}_{2}^{2} - \frac{1}{2} \text{mv}_{1}^{2}$
$= \frac{1}{2} m \left(\text{v}_{2}^{2} - \text{v}_{1}^{2}\right) = \frac{1}{2} \times \text{0.5} \left(\left(5 \times 2^{3 / 2}\right)^{2} - 0^{2}\right)$
$= \frac{1}{2} \times \text{0.5} \times 2 5 \times 8$
= 50 J