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Coefficients of friction between the flat bed of the truck and crate are $μ_{2} = 0.85$ and $μ_{h} = 0.8$ . The coefficient of kinetic friction between the truck tires and the road surface is 0.9. If the truck stops from an initial speed of $15 m / s$ with maximum braking (wheels skidding). Determine where on the bed the crane finally comes to rest. (Take, $g = 10 m / s^{2}$ )

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1.56 m

7.56m

6.56m

5.56m

answer is A.

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Detailed Solution

Since $μ$ , between crate and truck is less than 0.9. Therefore, retardation of truck and crane will be different.

Retardation of truck.

$a_{1} = 0.9 g = 9 m / s^{2}$

Retardation of crane,

$a_{2} = 0.8 g = 8 m / s^{2}$

Displacement of truck, before it finally stops is

$S_{1} = \frac{u^{2}}{2 a_{1}} = \frac{(15)^{2}}{2 \times 9} = 12.5 m$

and displacement of crane

$S_{2} = \frac{u^{2}}{2 a_{2}} = \frac{(15)^{2}}{2 \times 8} = 14.06 m ∴ S_{2} - S_{1} = 1.56 m$

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