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
Explanation:
We're asked to find the work done by the movers on a crate as it is moved
There are two forces acting on the crate here:
-
an applied force from the movers
-
a retarding kinetic friction force,
f_k , equal tof_k = mu_kn = mu_kmg
And we're given there is zero acceleration, so the net force is
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
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.