Question

At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. the coefficient of kinetic friction for the child sliding on the slide is 0.20. what is the magnitude of her acceleration during her sliding?

Answer 1

assuming metric units, metre, kilogram and seconds

Best approach: draw a free body diagram and identify forces acting on the child, which are:
gravity, which can be decomposed into normal and parallel (to slide) components
N=mg(cos(theta)) [pressing on slide surface]
F=mg(sin(theta)) [pushing child downwards, also cause for acceleration]
m=mass of child (in kg)
g=acceleration due to gravity = 9.81 m/s^2
theta=angle with horizontal = 42 degrees

Similarly, kinetic friction is slowing down the child, pushing against F, and equal to
Fr=mu*N=mu*mg(cos(theta))
mu=coefficient of kinetic friction = 0.2

The net force pushing child downwards along slide is therefore
Fnet=F-Fr
=mg(sin(theta))-mu*mg(cos(theta))
=mg(sin(theta)-mu*cos(theta)) [ assuming sin(theta)> mu*cos(theta) ]

From Newton's second law,
F=ma, or
a=F/m
=mg(sin(theta)-mu*cos(theta)) / m
= g(sin(theta)-mu*cos(theta)) [ m/s^2]

In case imperial units are used, g is approximately 32.2 feet/s^2.
and the answer will be in the same units [ft/s^2] since sin, cos and mu are pure numbers.

Answer 2

The magnitude of her acceleration during her sliding is 5.1m/s² .

Explanation:

The force exerted by the child = mass of the child × acceleration of the child = m × a = ma

but the child slides down at an angle = 42°

The force exerted by the child is inclined; there are 2 other components of child’s force that can be resolved in the plane of inclination, viz:

The horizontal component due to gravity = mgsin42°

The vertical component due to gravity and influenced by kinetic friction of the slide = µmgcos42°

Where g = acceleration due to gravity = 9. 8m/s², and

µ = coefficient of kinetic friction = 0.20

For equilibrium, the sum of forces towards the one direction = the sum of the forces in the opposite direction, thus:

ma + µmgcos42° = mgsin42°

Divide the equation all through by the mass of the child (m), and make the acceleration of the child (a) The subject of the formula, viz:

a = gsin42° — µgcos42°

= (9.8m × sin42°) — (9.8 × 0.2 × cos42°)

= 5.1m/s²