Answer:
The required work done is [tex]2555.448~J[/tex]
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '[tex]\mu_{k}[/tex]' be the coefficient of friction, then
[tex]f = \mu_{k} \times N = \mu_{k} \times Mg[/tex]
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
[tex]F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N[/tex]
So the work done (W) in moving the crate by a distance s = 10.6 m is
[tex]W = F \times s = 241.08~N \times 10.6~m = 2555.448 J[/tex]
