The coefficient of friction is 0.363
Explanation:
There are two forces acting on the scooter in the horizontal direction:
- The applied force, F = 1850 N, forward
- The frictional force, [tex]F_f[/tex], backward
Since the scooter is moving at constant speed, the acceleration is zero, so the net force acting on the scooter must be zero. Therefore we can write:
[tex]F-F_f = 0\\F_f = F = 1850 N[/tex]
The frictional force can be written as
[tex]F_f = \mu R[/tex] (1)
where
[tex]\mu[/tex] is the coefficient of friction
R is the normal reaction of the ground on the scooter
For a flat horizontal surface, there is equilibrium along the vertical direction, so the normal reaction is equal to the weight:
[tex]R = W = mg[/tex]
where
m = 520 kg is the mass
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting into (1),
[tex]F_f = \mu mg = 1850 N[/tex]
and solving for [tex]\mu[/tex],
[tex]\mu=\frac{F_f}{mg}=\frac{1850}{(520)(9.8)}=0.363[/tex]
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