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Fynjy0 [20]
3 years ago
8

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 i

s 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.
Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

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

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Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

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(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

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(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

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                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

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