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

A body of mass 0.5 kg travels in a straight line with velocity v = ax^

1.	A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2)where a = 5 m^(-1/2) s^(-1) . The work done by the net force during its displacement from x = 0 to x = 2 m is
Answer» `1.5` J 50J 10 J 100 J Solution :`v=ax^(3/2)` `m=0.5` kg `a =5 m^(-1/2) s^(-1)` Now `a_(0)=(dv)/(dt)` ` =v(dv)/(dx)` `=ax^(3/2) d/(dx)(ax^(3/2))` `=ax^(3/2) XX axx 3/2 XXX^(1/2)` `=3/2 a^(2)x^(2)` `= ax^(3/2)xxaxx 3/2 xxx^(1/2)` ` = 3/2a^(2)x^(2)` Now , `ma_(0)=m 3/2a^(2)x^(2)` Force , F SMALLER work done dW =Fdx Work done = `int_(xto0)^(x=2) F dx` ` = int_(0)^(2) 3/2 ma^(2)x^(2)dx ` `=3/2ma^(2)xx((x^(3))/3)_(0)^(2)` `=1/2ma^(2)xx8` `=1/2xx(0.5)xx(25)xx8 = 50 J `