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
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.