A mass of 1 kg is acted upon by a single force `F=(4hati+4hatj)N`. Under this force it is displaced from (0,0) to (1m,1m). If initially the speed of the particle was 2 `ms^(-1)`, its final speed should be

Options

A. `6 ms^(-1)`
B. `4.5 ms^(-1)`
C. `8 ms^(-1)`
D. `4 ms^(-1)`

Answer

B

Explanation

(b) Given, force, `F=(4hati+4hatj) N`
`r_(1)=0hati+0hatj`
`r_(2)=hati+hatj`
`r_(2)-r_(1)=hati+hatj`
Initial speed, `v_(1)=2 ms^(-1)`
From work energy theorem we have
`DeltaW=DeltaK`
` implies " " F.Delta r=(1)/(2)m(v_(2)^(2)-v_(1)^(2))`
`implies " "(4hati+4hatj).(hati+hatj)=(1)/(2)xx1(v_(2)^(2)-4)`
`implies " "4+4=(1)/(2)(v_(2)^(2)-4)`
`implies " " v_(2)^(2)=20 implies =5xx10=50 N`