physics
posted by josh on .
A 120kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34. What is the final speed of the crate?

find the work that went into the system: Force*distance
Now, find the frictional force on the next portion:
friction=mu*mg
Then find the work done on friction:
mu*mg*17
Finally, find the KE of the crate
KE crate=workinfrictionwork
1/2 mv^2=400*34mu(mg)17
v^2=800*34/mass 2*mu*g*17
solve for v
check my thinking.