The flatbed truck is traveling at the constant speed of 60 km/h up the 15‐percent grade when the 100‐kg crate which it carries is given a shove which imparts to it an initial relative velocity ˙ x = 3 m / s x˙=3 m/s toward the rear of the truck. If the crate slides a distance x = 2 m measured on the truck bed before coming to rest on the bed, compute the coefficient of kinetic friction μk between the crate and the truck bed.

μ = 0.309

Explanation:

coefficient of kinetic friction is defined as the ratio of two forces, friction force and the normal force acting on the object.

θ = arctan(15/100)= 8.531⁰

In the vertical direction:

N = mgcosθ = 100 *9.8 *cos(8.531) = 970N

law of conservation of energy implies

mgsinθ - μNx = 1/2m(v₂²-v₁²)

100*9.8*sin (8.531) - μ(970*2) = 1/2(100)(0²-3²)

150.6 - 1940μ = 450

- 1940μ = -600.6

μ = 0.309

the answer is b, have negligible mass

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