Question

Some sliding rocks approach the base of a hill with a speed of 13 m>s. The hill rises at 42° above the horizontal and has coefficients of kinetic friction and static friction of 0.43 and 0.61, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

Answer 1

Answer:

a )  - 8.32 m /s²

b ) 2.06  m /s² down the hill

Explanation:

Reaction of the hill surface = mgcos 42

a )  Friction force = μmgcos 42

where μ is coefficient of kinetic friction

Net downward force acting down the hill on rocks

= mgsin32 + μmgcos 42

acceleration

= -gsin32 - μgcos 42

=  - 9.8 x .5299 - .43 x 9.8 x cos42

=  - 5.19 - 3.13

= - 8.32 m /s²

b ) When it reaches the highest point , it becomes stationary . After that it tries to come down due to its weight.

Force acting downwards

= mgsin 42

5.19 m

Force of friction preventing it to slide down

= mgcos42 x .61 (  static friction will act on it  )

= 4.44m

The body will go down with  acceleration.

Net acceleration.  =

5.19 - 3.13  ( kinetic friction will act on it )

2.06  m /s² down the hill