Question

Find the work done by the force field F(x,y,z) = 7xi + 7yj + 4k on a particle that moves along a helix r(t) 2cos(t)i + 2sin(t)j + 6tk, 0 ≤ t ≤ 2π.

Answer 1

Field Force F(x,y,z) = 7x i + 7y j + 4 k

Helix path r(t) = 2cos(t)i + 2sin(t)j + 6tk = xi + yj + zk

By comparing above equations x = 2cos(t) , y = 2sin(t) and z = 6t

r(t)= (2Cos(t))i + (2Sin(t)j + 6tk

Apply derivative with respect to t on both sides.

dr(t) = -(2Sin(t))dt i + (2Cos(t))dt j + 6dt k

dr(t) = (-2Sin(t))dt i + (2Cos(t))dt j + 6dt k

The differential work is dot product of Force anf path length

dW = F•dr(t)

dW = ( 7xi + 7yj + 4k )•( (-2Sin(t))dt i + (2Cos(t))dt j + 6dt k )

= ( (7x)(-2Sin(t))dt i•i + (7y)(2Cos(t))dt j•j + (4*6)dt k•k )

Substitute i•i  = j•j = k•k = 1

= [-14xSin(t) + 14yCos(t) + 24]dt

Substitute x = 2cos(t) , y = 2sin(t)

[-14*2cos(t)Sin(t) + 14*2sin(t)yCos(t) + 24]dt

dW = [-28Cos(t)Sin(t) + 28Sin(t)Cos(t) + 24]dt

dW = 24dt

Now integrate with respect to t in the interval 0 to 2π

∫dW = ∫24dt

W = 24(2π-0)

W = 48π

W = 150.8 J

The work done is 150.8 Joules.