Question

A force of 10i^+20j^+5k^N is applied to a body due to which the body is displaced from the point 2,3,6 to 3,4,3 . Calculate the workdone where displacement measures in vector.

Answer 1

W = f.d
D = 3 ,4,3 -( 2,3,6)
D = 1i +1j-3k

W = 10i + 20j+ 5k ( I + j - 3k)
W = 10 + 20 - 15
W = 15 joule....


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Answer 2

Given:

Force = 10i + 20j + 5k  N

Displacement from (2,3,6) to (3,4,3)

To Find:

Work done = ?

Solution:

Displacement of Particle will be

⇒ ( 3i + 4 j + 3 k ) - ( 2i + 3 j + 6 k )

⇒ 1i +1j - 3k

Work done is given by

W = F. r =( 10i + 20j + 5k ) . ( 1i +1j - 3k )

W = 10 + 20 - 15

W = 15 joule

Hence, the workdone is 15 joule.