How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was .50?

1 Answer
Nathan L.
Aug 16, 2017

#W = 2300# #"J"#

Explanation:

We're asked to find the work done by the movers on a crate as it is moved #10.3# #"m"#.

There are two forces acting on the crate here:

  • an applied force from the movers

  • a retarding kinetic friction force, #f_k#, equal to #f_k = mu_kn = mu_kmg#

And we're given there is zero acceleration, so the net force is #0#.

The net force equation is thus

#sumF_x = F_"applied" - mu_kmg = 0#

And so

#ul(F_"applied" = mu_kmg#

We know:

  • #mu_k =0.50#

  • #m = 46.0# #"kg"#

  • #g = 9.81# #"m/s"^2#

Plugging these in:

#F_"applied" = 0.50(46.0color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(red)(ul(226color(white)(l)"N"#

The work done by this force is given by

#ul(W = Fs#

We're given the distance #s = 10.3# #"m"#, so we have

#W = (color(red)(226color(white)(l)"N"))(10.3color(white)(l)"m") = color(blue)(ulbar(|stackrel(" ")(" "2300color(white)(l)"J"" ")|)#

rounded to #2# significant figures.

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