Question

A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z) = z2i + 3xyj + 3y2k. Find the work done.

Answer 1

Answer:

[8i + 54j + 72k] joules per newton metre square.

Step-by-step explanation:

F(x,y,z) = z2i + 3xyj + 3y2k

F(3,0,0) = 0

F(3,3,1) = 1(2i) + 3(3)(3)j + 3(3)(2k)

= 2i + 27j + 18k

F(0,3,1) = 1(2i) + 3(0)(3)j + 3(3)(2k)

= 2i + 18k

The sum = [4i + 27j + 36k]

Since the particle moves back to the origin, we multiply the above Work Done by 2:

[8i + 54j + 72k]

And since work done is the product of Force and Distance, the S.I. unit is Joules/newton metre square

Answer 2

[8i + 54j + 72k] joules per newton metre square.

The work done by the force is the dot product of the force vector and the displacement vector of the particle.

Given:

F(x,y,z) = z2i + 3xyj + 3y2k

⇒F(3,0,0) = 0

⇒F(3,3,1) = 1(2i) + 3(3)(3)j + 3(3)(2k)  = 2i + 27j + 18k

⇒F(0,3,1) = 1(2i) + 3(0)(3)j + 3(3)(2k)  = 2i + 18k

The sum is [4i + 27j + 36k]

Since the particle moves back to the origin, we multiply the above Work Done by 2:

[8i + 54j + 72k]

And since work done is the product of Force and Distance, the S.I. unit is Joules/newton metre square.  

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