Question

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration 0.82 m/s2 directed down the ramp. The ramp makes an angle of 38° with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

Answer 1

Step-by-step explanation:

The given data is as follows.

a = 0.82
`m/s^(2)`, Angle (
`\theta`) =
`38^(o)`

So, in x-direction the force will act on the box. Hence, formula for normal force and x-component of the gravitational force is as follows.

`F_(n) = mg Cos (\theta)` ............. (1)

In y-direction,

`mg sin (\theta) - f = ma` .............. (2)

and, f =
`\mu_(k) F_(n)` ........... (3)

Now, we will use equations (1), and (3) in equation (2) as follows.

`mg sin (\theta) - (\mu_(k) mg cos (\theta)` = ma

Then,

`\mu_(k) = (a - g sin (\theta))/(g cos (\theta))`

=
`(9.8 sin (38) - 0.82)/(9.8 cos (38))`

=
`(2.9008 - 0.82)/(9.359)`

= 0.22

Thus, we can conclude that the coefficient of kinetic friction between the box and the ramp is 0.22.