A force #vecF = (3xy -5z)hatj + 4zhatk# is applied on a particle. The work done by the force when the particle moves from point #(0,0,0)# to point #(2,4,0)# along the path #y = x^2# is?

1 Answer
MetaPhysik
Feb 17, 2018

#W=48 [J]#

Explanation:

Use the definition of work,

#W = intvecf*dvecr#

Instead using unit vector notations, let's use coordinate to represent the vectors, i.e., #vec f = (0, 3xy -5z, 4z); and dvecr = (dx, dy, dz#)

#W = int_("(0,0,0)" )^("(2,4,0)")(0, 3xy -5z, 4z)*(dx, dy, dz#)#

#W = int_("(0,0,0)" )^("(2,4,0)") (3xy -5z)dy +4zdz#
#W= int_0^4 (3xy -5z)dy + int_0^0 4zdz #
#because# no displacement in z, z-component force contribute no work, hence

#W= [3/2xy^2-5zy]_0^4 + 0 #

#W= 24x-20z#
The above is the work done alone the path #y=x^2# on a z-plane.

At point (2,4, 0),

#W= 24x-20z= 24(2)-20(0)=48#

The units are not specified, hence the work done has not unit, otherwise it is assumed as #W=48J#

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