a 50kg box is pushed up a 30 degree plane which is 100m long. what is the speed of the box after being pushed the 100m? Mk=0.1
m*g = 50kg * 9.8N./kg = 490 N. = Wt. of
the box.
Fp = 490*sin30 = 245 N. = Force parallel
to the incine.
Fn = 490*cos30 = 424.4 N. = Force
perpendicular to the incline.
Fk = u*Fn = 0.1 * 424.4 = 42.44 N. =
Force of kinetic friction.
At bottom of incline:
KE + PE = mg*h - Fk*L
0 + PE = 490*100*sin30-42.44*100
PE = 24,500-4244 = 20,256 J.
At top of incline:
KE + PE = 20,256
KE + 0 = 20,256
0.5m*V^2 + 0 = 20,256
0.5*50*V^2 = 20,256
25V^2 = 20,256
V^2 = 810.24
V = 28.5 m/s.
Note: h = 100*sin30