A bear slides down an incline which is oriented at 17 degrees above horizontal. What is the coefficient of friction between the bear and the incline if the bear moves with a constant velocity?

Respuesta :

[tex]\mu = 0.306[/tex]

Explanation:

Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as

[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]

[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]

where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as

[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]

Combining these two equations together, we get

[tex]mg\sin17° = \mu(mg\cos17°)[/tex]

Solving for [tex]\mu, [/tex]

[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]