## 6304224 Group Title 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? 3 years ago 3 years ago

1. Owlfred

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2. 6304224

hi can u help me please

3. dumbcow

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) $50 \frac{km}{hr} = 50*\frac{1000}{3600}\frac{m}{s} = 13.889\frac{m}{s}$ $\frac{1}{2}k(2.2)^{2} = \frac{1}{2}(1200)(13.889)^{2}$ $k \approx 47,827.5$