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

Options

A. `1.5` J
B. 50 J
C. 10 J
D. 100 J

Answer

B

Explanation

`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 `