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?
1 answer:
When the car is moving, it has Kinetic Energy. As soon as the car entercount with the spring, its Kinetic Energy will transfer to Elastic Potential Energy. <span>KE = EPE </span> <span>1/2mv^2 = 1/2kx^2 </span> <span>change the speed into m/s </span> <span>50 km/hr = 13.9 m/s </span> <span>1/2(1200)(13.9)^2 = 1/2k(2.2)^2 </span> <span>now you do the math to solve for k. k= 115926/2.42 K= 47903.30</span>
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