The coefficient of sliding friction is 0.514
Explanation:
We start by writing the equations of motion of the elephant along the two directions, parallel and perpendicular, to the incline.
Along the parallel direction we have:
[tex]F- mg sin \theta - \mu_k R = ma[/tex] (1)
where :
F = 10,000 N is the force applied by Noah
[tex]mg sin \theta[/tex] is the component of the weight parallel to the incline, where:
m is the mass
g = 9.8 m/s^2 the acceleration of gravity
[tex]\theta=10^{\circ}[/tex] is the angle of incline
[tex]\mu_k R[/tex] is the force of friction, where:
[tex]\mu_k[/tex] is the coefficient of friction
R is the normal reaction
and a is the acceleration
Perpendicular direction:
[tex]R-mg cos \theta =0[/tex] (2)
where [tex]mg cos \theta[/tex] is the component of the weight perpendicular to the incline
From (2) we find
[tex]R=mg cos \theta[/tex]
And substituting into (1)
[tex]F-mg sin \theta - \mu_k mg cos \theta = ma[/tex]
We know that the elephant moves at constant speed, so the acceleration is zero:
a = 0
So the equation becomes
[tex]F-mg sin \theta - \mu_k mg cos \theta=0[/tex]
And we can re-arrange it to find the coefficient of friction:
[tex]F-mg sin \theta - \mu_k mg cos \theta=0\\\mu_k = \frac{F-m g sin \theta}{mg cos \theta}=\frac{10000-(1500)(9.8)(sin 10)}{(1500)(9.8)(cos 10)}=0.514[/tex]
Learn more about friction and inclined planes:
brainly.com/question/5884009
#LearnwithBrainly
