If under the action of a force (4i+j+3k)N, a particle moves from position r1=3i+2j-6k to position r2=14i+13j+9k, then the work done will be

Question

If under the action of a force (4i+j+3k)N, a particle moves from position r1=3i+2j-6k to position r2=14i+13j+9k, then the work done will be (A) 50 J (B) 75 J (C) 100 J (D) 175 J

Tardigrade Admin · Accepted Answer

Here, r 1=3 i +2 j -6 k and r 2 =14 i +13 j +9 k So, displacement, ( r 2 - r 1 )= (14 i +13 j +9 k )-(3 i +2 j -6 k ) =11 i +11 j +15 k Hence, work done = F ⋅ s =(4 i + j +3 k ) ⋅(11 i +11 j +15 k ) =44+11+45=100 J

Q. If under the action of a force $(4 i + j + 3 k) N,$ a particle moves from position $r_{1} = 3 i + 2 j - 6 k$ to position $r_{2} = 14 i + 13 j + 9 k,$ then the work done will be

50 J

75 J

100 J

175 J

Here, $r_{1} = 3 i + 2 j - 6 k$
and $r_{2} = 14 i + 13 j + 9 k$
So, displacement, $(r_{2} - r_{1}) =$
$(14 i + 13 j + 9 k) - (3 i + 2 j - 6 k)$
$= 11 i + 11 j + 15 k$
Hence, work done
$= F \cdot s$
$= (4 i + j + 3 k) \cdot (11 i + 11 j + 15 k)$
$= 44 + 11 + 45 = 100 J$

Q. If under the action of a force (4i+j+3k)N, a particle moves from position r1​=3i+2j−6k to position r2​=14i+13j+9k, then the work done will be