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Sagot :
Answer:
The coefficient of friction between the crate and the truck that is required to keep the crate from sliding is 0.23
Explanation:The initial veocity of the truck, u = 13.6 m/s
The distance covered, s = 41.0 m
The driver applies the brakes and the truck is brought to a halt
The final velocity = 0 m/s
Calculate the deceleration by using the equation of motion below
[tex]\begin{gathered} v^2=u^2+2as \\ 0^2=13.6^2+2a(41) \\ 82a=-13.6^2 \\ 82a=-184.96 \\ a=-\frac{184.96}{82} \\ a=-2.26m/s^2 \end{gathered}[/tex]Calculate the coefficient of friction using the relationship below
[tex]\begin{gathered} a=-\mu g \\ -2.26=-9.8\mu \\ \mu=-\frac{2.26}{-9.8} \\ \mu=0.23 \end{gathered}[/tex]The coefficient of friction between the crate and the truck that is required to keep the crate from sliding is 0.23
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