Question

a body is subjected to a constant force given by F= -i^+ 2j+3k^. what is work done by this force in moving the body through a distance of 4m along the z acis and then 3 m along y axis​

Answer 1

Answer:

Work done is equal to force × displacement .so vector × vector cross product.I did this in matrix way check the photo for solution

Answer 2

The work done is 18 Joule.

Explanation:

The work done on the body is given as

W = F.d

Here F is the force given to the body and d is its displacement after subjected to a force to the body.

As body moves 4 m along z axis and 3 m along y axis,

In vector notation,

d = (4 m) k + (3 m) j.

given F = (-i +2j + 3 k ) N.

The dot product is given as

i.i = j.j = k.k = 1

and  i.j = j.k = k.i =0.

Substitute the given values, we get

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

$W=12(k.k)+6(j.j)$

W = 18 J.

Thus, the work done is 18 Joule.

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