Question

A force of F=2xi+2j+3z^2k N is acting on a particle. Find the work done by this force in displacing the body from (1,2,3)m to (3,6,1)m.

Answer 1

final position - initial = 2i+4j-2k m
workdone = f.s 
  =(2xi x 2i ) + (2j x 4j ) + (3z^2k x -2k )
 = 4x+8-6z^2 joules

Answer 2

Dear Student,

◆ Answer -

W = 6 J

● Explanation -

Position vectors for two points are -

a = i + 2j + 3k

b = 3i + 6j + k

Displacement of the body is -

s = b - a

s = (3i + 6j + k) - (i + 2j + 3k)

s = 2i + 4j - 2k

Work done is calculated by -

W = F.s

W = (2i + 2j + 3k).(2i + 4j - 2k)

W = 4 + 8 - 6

W = 6 J

Hence, work done by the force in displacing the body is 6 J.

Thanks dear. Hope this helps you...