Question
The man pushes/pulls with a force of 200N. The child and the slide combo has a mass of 30 kg and coefficient of friction is 0.15. The angle between the slide and the ground is 30 degrees. Find the acceleration of the child and what is the frictional forces opposing his efforts
Answers
M*g = 30 * 9.8 = 294 N.
Fp = 294*sin30 = 147 N. = Force parallel with slide.
Fn = 294*cos30 = 254.6 N. = Force perpendicular to the ride = Normal.
Fk = u*Fn = 0.15 * 254.6 = 38.2 N. = Force of kinetic friction.
Fap = 200 N. = Force applied.
Fap-Fp-Fk = M*a.
200-147-38.2 = 30*a,
30a = 14.8, a = 0.493 m/s^2.
Fp = 294*sin30 = 147 N. = Force parallel with slide.
Fn = 294*cos30 = 254.6 N. = Force perpendicular to the ride = Normal.
Fk = u*Fn = 0.15 * 254.6 = 38.2 N. = Force of kinetic friction.
Fap = 200 N. = Force applied.
Fap-Fp-Fk = M*a.
200-147-38.2 = 30*a,
30a = 14.8, a = 0.493 m/s^2.