Question

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a) Determine the work done by the force F = (xy + 3z)i + (2y2-x2)i + (z-2y)k in
taking a particle from x=0 to x = 1 along a curve defined by the equations:
x2 = 2y;
2x3 = 32​

Answer 1

GIVEN :

F = (xy + 3z)i + (2y2-x2)i + (z-2y)k

Particle moves from x=0 to x = 1 along a curve defined by the equations:

x2 = 2y;

2x3 = 32

SOLUTION :

◆Work done  W=∫F. dl          

where F - Force and dl - displacement and are vectors.

◆Since the particle is taken from. x= 0 to 1, in x direction,

dl= dx i^

◆(i^ , j^, k^ - unit vector along x ,y and z respectively.)

◆As, F = (xy + 3z)i^ + (2y2-x2)j^ + (z-2y)k^

W = ∫F.dl

◆Substituting values,

= [(xy + 3z)i^ + (2y2-x2)j^ + (z-2y)k^ ]. dx i^

◆W=( xy + 3z).dx --(1)

◆Given,

x^2 = 2y , y = x^2 /2 ,

2x^3 = 3z , z = 2x^3 / 3

◆Substituting in equation (1)

F.dl= x^3/2 + 2x^3 = 5x³/2

◆Integrating along x from 0 to 1

W = ∫F.dl = 5/8.

ANSWER :

W = ∫F.dl = 5/8.