physics #2?

Question

The carton lies on a plane tilted at an angle of 22.0 degrees to the horizontal with uk= 0.12.

a.) Determine the acceleration of the carton as it slides down the plane.

b.) IF the carton starts from rest 9.30 m up the plane from its base, what will be the carton's speed when it reaches the bottom of the incline.

Answer 1

First, decompose the forces due to gravity on the carton:

Perpendicular = cos(22)*m*g
Parallel = sin(22)*m*g

Assuming the carton is sliding, compute the force due to friction:

f=Cos(22)*m*g*uk

so the acceleration of the carton is
(sin(22)-Cos(22)*.12)*m*g/m
or (sin(22)-Cos(22)*.12)*9.8
=2.58m/s^2

To calculate the velocity
average velocity is 1/2a*t
and distance is
1/2a*t^2
we know that the distance is 9.3m
so we can compute t as
sqrt(9.3*2/2.58)
= 2.68s
the velocity at the end is
2.68*2.58
=6.9m/s

j