Question

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 = 2m ?

Answer 1

it is given that, $v=ax^{3/2}$ where a = 5m½/s

at (x = 0 ) , kinetic energy = 1/2 mv²
= 1/2 × 0.5 × {a(0)^3/2}²
= 0

at (x = 2m) , kinetic energy = 1/2mV²
= 1/2 × 0.5 × {a(2)^3/2}²
= 1/2 × 0.5 × a² × (2)³
= 1/2 × 0.5 × 5² × 8
= 12.5 × 4 = 50 J

workdone = change in Kinetic energy
= kinetic energy at (x = 2m) - kinetic energy at(x = 0)
= 50J - 0

= 50J