A smooth track in the form of a quarter circle of radius `6 m` lies in the vertical plane. A particle moves from `P_(1)` to `P_(2)` under the action of forces `vec F_(1),vecF_(2)` and `vec F_(3)` Force `vec F_(1)` is 20N, Force `vec F_(2)` is always `30 N` in magnitude. Force `vec F_(3)` always acts tangentiallly to the track and is of magnitude `15 N`. Select the correct alternative (s) :
.

Options

A. Work done by `vec F_(1)` is `120 J`
B. Work done by `vec F_(2)` is `180 J`
C. Work done by `vec F_(3)` is `45 pi J`
D. `vec F_(1)` is conservative in nature.

Answer

B, C, D

Explanation

Work done by `vec F_(1)` is
`W_(1)=int_(P_(1))^(P_(2)) F_(1)cos theta ds`
Here, `ds=(6)d(-2 theta)=-12 d theta` and `F_(1)=20 N`
`:. W_(1)=-240 int_(x//4)^(0) cos theta d theta = 240 sin (pi)/(4) = 120 sqrt(2)J`
`vec F_(1)` is conservative because it is always directed towards a fixed point `P_(2)`.
`W_(2)=F_(2)(OP_(2))=(30)(6)=180J` and
`W_(3)=vecF_(1) int_(0)^(6(pi//2))F_(3)ds =int_(0)^(3pi) 15ds =[15x]_(0)^(3pi)= 45 pi`.