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dezoksy [38]
3 years ago
12

Please solve it asap ty!

Physics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

thanks for the five points

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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
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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:

-F_{f} + F = 0      

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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:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

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.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

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