Question

Find the work done by the force field F(x,y,z)=6xi+6yj+6k on a particle that moves along the helix r(t)=3 cos(t)i+3sin(t)j+2 tk,0≤t≤2π.

Answer 1

Answer:

the work done by the force field = 24 π

Step-by-step explanation:

From the information given:

r(t) = 3 cos (t)i + 3 sin (t) j + 2 tk

= xi + yj + zk

∴

x = 3 cos (t)

y =  3 sin (t)

z = 2t

dr = (-3 sin (t)i + 3 cos (t) j + 2 k ) dt

Also F(x,y,z) = 6xi + 6yj + 6k

∴  F(t) = 18 cos (t) i + 18 sin (t) j +6 k

Workdone = 0 to 2π ∫ F(t) dr

$\mathbf{= \int \limits ^{2 \pi} _{0} (18 cos (t) i + 18 sin (t) j +6k)(-3 sin (t)i+3cos (t) j +2k)\ dt}$

$\mathbf{= \int \limits ^{2 \pi} _{0} (-54 \ cos (t).sin(t) + 54 \ sin (t).cos (t) + 12 ) \ dt}$

$\mathbf{= \int \limits ^{2 \pi} _{0} 12 \ dt}$

$\mathbf{= 12 \times 2 \pi}$

= 24 π