Net force downward= 30*9.8*cos40 - 30*9.8*.2
acceleration= net force downward / mass
a 30 kg box is placed on a 40 degree inclined plane the coefficient of friction is 0.2 at what rate does the box accelerate down the incline
2 answers
Mg = 30*9.8 = 294 N.
294*sin40 = 189 N. = Force parallel with plane.
Fn = 294*cos40 = 225 N. = Normal force.
u*Fn = 0.2 * 225 = 45 N. = Force of kinetic friction.
Mg-u*Fn = M*a
294-45 = 30*a
a = ___ m/s^2.
294*sin40 = 189 N. = Force parallel with plane.
Fn = 294*cos40 = 225 N. = Normal force.
u*Fn = 0.2 * 225 = 45 N. = Force of kinetic friction.
Mg-u*Fn = M*a
294-45 = 30*a
a = ___ m/s^2.
2 answers