The coefficient of sliding friction is 0.227
Explanation:
We can solve the problem by applying Newton's second law:
[tex]\sum F = ma[/tex]
where
[tex]\sum F[/tex] is the net force acting on the truck
m is the mass of the truck
a is its acceleration
We are told that the truck is pulled at constant speed: this means that its acceleration is zero, so
a = 0
and the equation becomes
[tex]\sum F = 0[/tex]
There are two forces acting on the truck:
- The horizontal pushing force, F = 400 N
- The frictional force, whose magnitude is given by [tex]F_f = \mu R[/tex], where [tex]\mu[/tex] is the coefficient of sliding friction and R is the normal reaction exerted by the floor on the truck.
For a horizontal surface, the normal reaction is equal to the weight, so:
R = W = 1760 N (weight of the truck)
So the frictional force becomes
[tex]F_f = \mu W[/tex]
And the net force therefore is
[tex]\sum F = F - \mu W = 0[/tex]
from which we can find the coefficient of friction:
[tex]\mu = \frac{F}{W}=\frac{400}{1760}=0.227[/tex]
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