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lions [1.4K]
3 years ago
13

(II) A 1200-kg car moving on a horizontal surface has speed v = 85 km/h when it strikes a horizontal coiled spring and is brough

t to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
Physics
1 answer:
Pie3 years ago
7 0

Answer:

k = 138440 N/m

Explanation:

given,

Mass of the car = 1200 Kg

speed of the car = 85 km/h

                            = 85 x 0.278 = 23.63 m/s

KE of the car

KE = \dfrac{1}{2} mv^2

Using Spring Energy Formula

KE_{spring}=\dfrac{1}{2}kx^2

Using conservation of energy

\dfrac{1}{2} mv^2=\dfrac{1}{2}kx^2

k = \dfrac{mv^2}{x^2}

k = \dfrac{1200\times 23.63^2}{2.2^2}

    k = 138440 N/m

Spring stiffness constant of the spring is equal to k = 138440 N/m

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