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

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide=$42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to  $a = 5.1\ m/s^2$