There are several ways to do this. Let's try conservation of energy. The loss of potential coming down (M g H)is equal to the gain in kinetic energy, (1/2) M V^2, PLUS work done against friction while sliding. The length of the slide is H/sin 28 = 7.46m, where H = 3.5 m is the vertical height. The friction force is M g cos28*uk. The friction work done is therefore
W = M g (H/sin 28)* cos28 *uk
= M g uk/tan28
M g H = M g H uk/tan 28 + (1/2)M V^2
Note that the M's cancel. That's good, because they did not tell you what the mass was.
V^2 = 2 g H [1 - uk/tan28]
I need help with this physics problem: A child slides down a playground slide; the coefficient of kinetic friction is uk=0.25, and angle that the slide makes with the horizontal is 28.0 degree. She begins from rest and slides through a vertical height of 3.5 m. With what speed does she exit the slide?
1 answer
1 answer