Consider a particle on which constant forces ` F _1 = hati +2 hatj + 3hatk ` N and ` F_2 = 4 hati - 5 hatj - 2 hatk ` act together resulting in a displacement from position `r_1 = 20 hati + 15 hatj ` cm ` to ` r_2 = 7 hatk ` cm . the total work done on the particle is

Options

A. `-0.48 J`
B. `+0.48 J`
C. `-4.8 J`
D. `+4.8 J`

Answer

A

Explanation

Given ,
` F_1 =( hati + 2 hatj + 3hatk ) N`
`F_2 =( 4 hati - 5 hatj - 2 hatk )N`
` r_1 = 20 hati + 15 hatj `cm
` r_2 = 7 hatk ` cm
total force on particle , `F = F_1 +F_2`
` = hati + 2hatj + 3hatk + 4 hati - 5 hatj - 2 hatk `
` = ( 5 hati - 3 hatj + hatk )N`
Displacement of particle `s= r_2 - r_1`
` = 7 hatk - 20 hati - 15 hatj`
`= ( -20 hati - 15hatj + 7 hatk ) cm `
we know that ,
Work `( omega ) = F.s `
`=( 5 hati - 3 hatj + hatk)(-20 hati - 15 hatj + 7 hatk ) xx 10 ^(-2)`
`=- 0.48 J`