Question #e7c74

1 Answer
Aug 19, 2017

#"net force" = 0#

#"friction force" = 100# #"N"# (directed opposite to applied force)

Explanation:

We're asked to find

  • the net force acting on the crate

  • the magnitude of the friction force acting on the crate

#" "#

  • Net force

We're given that #100# #"N"# of applied horizontal force is necessary so that the crate moves at a constant speed across the floor.

Since it is moving at a constant speed, it has zero acceleration, and according to Newton's second law, the net force is zero:

#F_"net" = ma#

#color(red)(ulbar(|stackrel(" ")(" "F_"net" = m(0) = ul(0)" ")|)#

#" "#

  • Friction force

Since we know the net force is zero, then the vector sum of the applied force and the friction force is equal to #0#:

#F_"net" = F_"applied" - F_"friction" = 0#

We're given that the applied force is #100# #"N"#, so we have

#100color(white)(l)"N" - F_"friction" = 0#

And so

#color(blue)(ulbar(|stackrel( "")(" "F_"friction" = 100color(white)(l)"N"" ")|)#

in the opposite direction of the applied force.