Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
[tex] -F_{f} + F = 0 [/tex]
[tex] -F_{f} + ma = 0 [/tex]
[tex] \mu mg = ma [/tex]
[tex] \mu = \frac{a}{g} [/tex]
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]
Where:
d: is the distance traveled = 46.1 m
[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)
[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s
[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
[tex] \mu = \frac{a}{g} [/tex]
[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]
[tex] \mu = 0.35 [/tex]
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
