Question

(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 brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Answer 1

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