Question

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

Answer 1

Answer:

Work done by net force during displacement is 50 J.

Explanation:

Mass of the body,"m" = 0.5kg

Velocity by the equation ,"v"  $=ax^{\frac{3}{2}}  where a=5m^{\frac{1}{2}}s^{-1}$

Initial velocity, "u" (at x=0) = 0

Final Velocity, "v" (at x=2m) = $10\sqrt{2}  m/s$

Formula:

            Work done (W)= Change in Kinetic Energy

$=(\frac{1}{2})m(x^{2}-x^{2})\\=(\frac{1}{2})0.5[(10\sqrt{2})^{2}\\=(\frac{1}{2}))0.5*10*10*2\\ = 50J$

Answer 2

generally, There are two methods for solving this question. first one is already mentioned in 1st answer. second one , I want to put here.

velocity, v = $ax^{3/2}$

differentiating with respect to time,

dv/dt = 3/2 a√x (dx/dt)

or, acceleration, A = 3/2 a√x × 3/2 a^{3/2}

= 3/2 a² x²

so, Force, F = mA = 3/2 ma²x²

now, workdone = ∫ F.dx

= 3/2 ma² ∫x² dx

= 3/2 ma² x³/3

= 1/2 ma²x³

putting x = 2 , m = 0.5, a = 5 and x = 2

then, Workdone = 1/2 × 0.5 × (5)² × (2)³

= 50J

hence, workdone by the net force during its displacement from x = 0 to x = 2m.

[ note :- this method seems little hard. but it will help you understand physics more conceptually]