Question

  A child of mass 45 kg goes down a playground slide of length 3.0 m that is inclined at an angle of
30° with respect to the horizontal. Find the speed of the child at the bottom given that the coefficient of kinetic friction between the child and the slide is 0.15


My main problem is reasoning through this type of problem.
I solved for the components x-component=2.81m and y-component= 1.02 m.

Then solved for fg=mg and its 441N. I think that this has to be equal to the normal force since the child isnt moving up and down.

so im left with the formula Fk=u*Fn and Fk= 66.15N. i tried dividing by 9.8m/s2 but i dont think thats the answer. Can someone guide me through this?

Answer 1

The net force is the force parallel minus the friction force.
Force parallel = 45 * 9.8 * sin 30 = 220.5 N
Ff = 0.15 * 45 * 9.8 * cos 30 = 66.15 * cos 30

Net force = 220.5 – 66.15 * cos 30
a = (220.5 – 66.15 * cos 30) ÷ 45 = 4.9 – 1.47 * cos 30
This is approximately 3.27 m/s^2. Use the following equation to determine the final velocity.

vf^2 = vi^2 + 2 * a * d, vi = 0
vf^2 = 2 * (4.9 – 1.47 * cos 30) * 3
vf = √(29.4 – 8.82 * cos 30)
This is approximately 4.7 m/s.