Question

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.

Answer 1

Answer:

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

Answer 2

The coefficient of kinetic friction ( μk ) = 0.309

Given data :

Speed of truck = 60 km/h

initial relative velocity ( x ) = 3 m/s

Distance travelled by crate = 2 m

Determine the coefficient of kinetic friction between the crate and truck bed

θ = arctan ( $(15/100)=$ 8.531°

considering movement in the vertical direction

N = mg*cos θ

   = 100*9.8 *cos(8.531) = 970 N

next step : determine the coefficient of kinetic friction

applying the law of conservation of energy

mg *sinθ - μxN = 1/2 *m(v₂²-v₁²)

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

∴ μx = 0.309

Hence we can conclude that The coefficient of kinetic friction ( μk ) = 0.309

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