A child is sliding down a slide, that makes 41.6o with respect to the ground. The coefficient of friction between the child's clothes and the slide is 0.4.
What is the child's acceleration?
Apply,
Fnet = m*g*sin(41.6) - mue_k*m*g*cos(41.6) = 0
m*a = m*g*sin(41.6) - mue_k*m*g*cos(41.6)
a = g*sin(41.6) - mue_k*g*cos(41.6)
= 9.8*sin(41.6) - 0.4*9.8*cos(41.6)
= 3.57 m/s^2
Given,
k = 0.4
theta = 41.6 degree
a =to be calculated
Let us consider "m" be the child's mass.
F(friction) = kN =
kmgsin(theta)
where normal force N = mg sin(theta)
Also, F = mxa, equating both the eqn we get,
mxa = kmgsin(theta),
which gives us
a = k xg sin (theta) =
0.4 x 9.8 x sin(41.6) = 0.4 x 9.8 x 0.663 = 2.598
m/sec2
Hence, a = 2.598 m/sec2
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A
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