A particle of mass m moves from A to C under the action of force `vecF = 2xyhati + y^2 hatj` , along different paths as shown in figure.

Options

A. Force is conservative in nature
B. Work done by force is `1/3` when particle moves along path ABC
C. Force is non conservative in nature and work done along path AC is 1 J
D. Work done along path ADC is `14/5 J`

Answer

a;;b

Explanation

`W = int barF d bars = int(2 xy hati + y^2 hatj) cdot (dx hati + dy hatj) = int 2xy dx + y^2 dy`
Along AC : As `x = y` and `dx = dy implies W_(AC) = int_(0)^(1) 3x^2 dx = 1 J`
Along AD : `x = 0 implies W_(AD) = int_(0)^(1) y^2 dy = 1//3 J , " Along " DC : y = 1 implies dy = 0`
`implies W_(DC) = int_(0)^(1) 2xdx = 1J`
`implies W_(ADC) = W_(AD) + W_(DC) = 1//3 + 1 = 4//3 J`
Along AB : `y = 0 implies W_(AB) = 0`
Along BC : `x = 1 implies dx = 0 implies W_(BC) = int_(0)^(1) Y^2dy = 1//3 J implies W_(ABC) = W_(AB) + W_(BC)F = 0 + 1//3 = 1//3 J`.