A body of mass 0.5 kg travels in a straight line with velocity `v= kx^(3//2)` where `k=5m^(-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

D

Explanation

Given , m = 0.5 kg `v = ax^(3//2) , a = 5 m ^(-1//2) s^(-1)`
Since work done
` w = int F dx = int m (dv)/(dt) dx = int m dv v `
or `W = m int c dv " " ` …(i)
Since ` v = ax^(3//2)= 5x^(3//2) `
At x = 0 , v = 0
Atx = 2 , ` v = 5 (2) 3/2 = 10 sqrt(2)`
integrating equation (i) with in above limits , we get
`W = m int_(0)^(10sqrt(2)) vdv = 0.5 [(v^(2))/2 ] _(0)^(10sqrt(2))`
` = (0.5)/2 xx 100 xx 2 = 50 J `.