how did you get a net force of 3N?
net force=283-mg*.357 ?
Is this thing going downhill? with a horizontal force?
I am lost on what is happening here.
A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.845 m/s encounters a rough horizontal surface of length scripted l = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.357 and he exerts a constant horizontal force of 283 N on the crate, find the following.
(c) Find the speed of the crate when it reaches the end of the rough surface.
I already found the net force to be 3N and the work done on the crate while its on the rough surface to be -25.4 J.
To find the final speed I attempted to use work=mass*distance*acceleration to solve for a and I got a=.424
Then I used v(final)^2=v(inital)^2+2a*deltax
my answer for this was incorrect.
Can someone please explain why this was the incorrect approach and how I should have gone about solving the problem? Thanks!
1 answer