A 20.0 kg box slides down a 12.0 m long incline at an angle of 3.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is
3 answers
593j
its wrong!!!!!!
M*g = 20 * 9.8 = 196 N. = Wt. of box.
Fp = 196*sin3 = 10.26 N. = Force parallel to the incline.
Fn = 196*Cos3 = 195.7 N. = Force perpendicular to the incline.
Fk = u*Fn = 0.1 * 195.7 = 19.57 N. = Force of kinetic friction.
a = (Fap-Fp-Fk)/M = (50-10.26-19.57)/20 =1.01 m/s^2
V^2 = Vo^2 + 2a*d = 0 + 2.02*12 = 24.24
V = 4.92 m/s.
KE = 0.5*20*4.92^2 = 242 J.
Fp = 196*sin3 = 10.26 N. = Force parallel to the incline.
Fn = 196*Cos3 = 195.7 N. = Force perpendicular to the incline.
Fk = u*Fn = 0.1 * 195.7 = 19.57 N. = Force of kinetic friction.
a = (Fap-Fp-Fk)/M = (50-10.26-19.57)/20 =1.01 m/s^2
V^2 = Vo^2 + 2a*d = 0 + 2.02*12 = 24.24
V = 4.92 m/s.
KE = 0.5*20*4.92^2 = 242 J.
3 answers