Question

A 1200 kg car rolling on a horizontal surface has speed v = 50 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

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Answer 2

hi can u help me please

Answer 3

the potential energy of the spring comes from the spring equation
\[PE = \frac{1}{2}kx^{2}\]
x is the displacement of the spring, k is the constant
the kinetic energy of the car comes from the equation
\[KE=\frac{1}{2}mv^{2}\]
m is mass of car, v is velocity in meters per second
To solve for constant k, set the KE of car equal to PE of spring
(for the car to come to rest then all of its KE is transferred to the spring)
\[50 \frac{km}{hr} = 50*\frac{1000}{3600}\frac{m}{s} = 13.889\frac{m}{s}\]
\[\frac{1}{2}k(2.2)^{2} = \frac{1}{2}(1200)(13.889)^{2}\]
\[k \approx 47,827.5\]