A force `vecF=(3xN)hati+(4N)hatj`, with x in meter, acts on a particle, changing only the kinetic energy of the particle. How mcuh work is done on the particle as it moves from coordinates `(2m, 3m, 5m)` to `(3m, 0m, 6m)`? Does the speed of the particle increase, decrease, or remain the same?

Options

A. `-7 J`
B. zero
C. `+ 7J`
D. `+ 19 J`

Answer

C

Explanation

`vecF=3x^(2) hat i+4 hat j`
` vecr =x hat i+y hati because vec(dr)=dx hat i +dy hat j`
Work done,
`W=int vecF.dvecr`
`=underset(((2,3)))overset(((3,0)))int(3x^(3)hat i=4 hat j).(dxhat i+dy hat j)`
`=underset(((2,3)))overset(((3,0)))int3x^(2)dx+4dy = underset(((2,0)))overset(((3,0)))intd(x^(3) +4y)`
`=[x^(3)+4y ]_(((2,3)))^(((3,0)))=3^(2)+4xx0-(2^(3)+4xx3)`
` =27 +0 -(8=12) =27 -20 =+7 j`
According to work -energy theorem,
Change in the kinetic energy=Work done, `W=+7J`.