M*g = 92 * 9.8 = 901.6 N. = Wt. of crate
= Normal force, Fn.
Fk = u*Fn = 0.357 * 901.6 = 321.9 N. =
Force of kinetic friction.
a. Fex-Fk = 289 - 321.9 = -32.9 N.
The Force of friction is greater than the exerted force.
A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encounters a rough horizontal surface of length 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 289 N on the crate, find the following.
a.) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
Magnitude____N
-I got 32.8 N
b.) Find the net done on the crate while it is on the rough surface. ____J
-I got this wrong for some reason.
Wnet=(32.8N)(cos180)(0.65)
Wnet=21.36 J???
c.) (c) Find the speed of the crate when it reaches the end of the rough surface. ____m/s
- I got 1.26 m/s which is apparently wrong also
Wnet=1/2mvf^2-1/2mvi^2
1 answer