A 1370 kg car rolling on a horizontal surface has a speed of 20 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 5.5 m. What is the spring constant (in N/m) of the spring? (Ignore nonconservative forces such as friction.)

Answer 1

Loss in KE = Gain in elastic PE of the spring.
(1/2)mV^2 = (1/2)kx^2
find K from this equation. Remember to convert the units of V to m/s