The constant forces f1 = 8 + 29 + 32 n and f2 = 48 - 59 - 22 n act together on a particle during a displacement from the point a (20, 15,0)m to the point b (0,07) m. what is the work done on the particle? the work done is given by f. f, where is the resultant force (here f = fi + f2) and is the displacement.

-600 J

Explanation:

F₁ = 8i +29 j + 32k

F₂ = 48 i - 59 j - 22 k

F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k

F = 56i - 30 j + 10 k

displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k

d = - 20 i - 15 j + 7 k  

Work Done = F dot product d

F . d = - 56 x 20 - 30 x - 15 + 10 x 7

=  - 1120 +450 + 70

= -600 J

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

double the distance = divide strength by 4