Question

A child is sliding down a slide, that makes 41.6^o 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?

Answer 1

a= miu*g*sin thetha
=0.4*9.8*sin(41.6)
= 2.6 m/s^2

Answer 2

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

Answer 3

Given,

$\mu$ k = 0.4

theta = 41.6 degree

a =to be calculated

Let us consider "m" be the child's mass.

F(friction) = $\mu$ kN = $\mu$ kmgsin(theta) where normal force N = mg sin(theta)

Also, F = mxa, equating both the eqn we get,

mxa = $\mu$ _kmgsin(theta), which gives us

a = $\mu$ 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