Question #f2bd1

A force is said to do work if, when acting on a body, there is a displacement of the point of application in the direction of the force.

The work done by a constant force of magnitude #F# on a point that moves a displacement #s# in the direction of the force is their product and can be expressed as follows:

#W=F*s#

So, in this case,

Work done,
#W=(567)*(20)#
#W=11340# joules

I think the weight of the body is perpendicular to the displacement in this case. So, as the force applied along the displacement or at any other angle other than #90^0# is not given, we can't calculate the work done. If there is no external force acting on the crate and the crate moved 20m just due to inertia, the work done is zero. All I can say is the question lacks details.

Plus, the dot product of F and S is #Fscostheta#. Here the weight of the crate acts downwards and I assume the question meant displacement is horizontal. So, we have #Fscos90^0# which is 0 as #cos90^0# is 0.

The question would have been clearer if the direction of applied force and its magnitude were given or the direction of displacement was given.