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?

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.

godwinihagh godwinihagh · Answer 2 · 2020-02-10T09:54:48+01:00

Answer:

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

Step-by-step 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