The work done in moving a particle in the force field F = 3x² i + (2xz - 9y) j+ zk along the line joining (0 , 0, 0) to (2 ,1, 3) is

a)
0

b)
16

c)
24

d)
18

Correct answer is option 'B'. Can you explain this answer?

Question

The work done in moving a particle in the force field F = 3x² i + (2xz - 9y) j+ zk along the line joining (0 , 0, 0) to (2 ,1, 3) is

a)
0
b)
16
c)
24
d)
18

Correct answer is option 'B'. Can you explain this answer?

Answer

Actually I think the answer is 12 and the jth component has 9y so it will have a 9t and will cancel with the other component when done dot product so we will remain with integral 0 to 1 36t^2dt

Answer

To calculate the work done in moving a particle in a force field, we need to integrate the dot product of the force and displacement vectors.

In this case, the force field is F = 3x, where x is the position vector of the particle.

Let's assume the particle moves from position A to position B. The displacement vector is given by Δx = B - A.

The work done is given by the integral of the dot product of the force and displacement vectors:

Work = ∫ (F · Δx)

Since F = 3x, we have:

Work = ∫ (3x · Δx)

Let's assume the particle moves along the x-axis from x = a to x = b. In this case, Δx = (b - a) i, where i is the unit vector along the x-axis.

So, the work becomes:

Work = ∫ (3x · (b - a) i)

= ∫ (3x(b - a)) dx

= 3(b - a) ∫ x dx

= 3(b - a) [(1/2) x^2]

= (3/2)(b - a) (x^2)

To evaluate the integral, we substitute the limits of integration:

Work = (3/2)(b - a) [(b^2) - (a^2)]

= (3/2)(b - a)(b^2 - a^2)

= (3/2)(b - a)(b + a)(b - a)

= (3/2)(b - a)^2(b + a)

Therefore, the work done in moving the particle in the force field F = 3x from position A to position B is (3/2)(b - a)^2(b + a).

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