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A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding?

Answer :

Answer:0.3

Explanation:

Given

velocity of car=15 m/s

truck brought to halt in a distance of 38 m

We know

[tex]v^2-u^2=2as[/tex]

Final velocity (v)=0

[tex]0-(15)^2=2(a)(38)[/tex]

[tex]a=\frac{-225}{76}[/tex]

[tex]a=-2.96 m/s^2[/tex]  (deceleration)

Therefore minimum coefficient of friction \mu will be

[tex]\mu \times g=a[/tex]

[tex]\mu =\frac{a}{g}[/tex]

[tex]\mu =\frac{2.96}{9.8}=0.302[/tex]

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