A 950 kg car rolling on a horizontal surface has speed v = 45 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?
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45 km/h=12.5 m/s
conservation of energy
PE=0.5*m*v^2=0.5*k*x^2
k=m*v^2/x^2
=0.5*950*12.5^2/2.2^2
=15334.45 N/m
=15.33 kN/m
conservation of energy
PE=0.5*m*v^2=0.5*k*x^2
k=m*v^2/x^2
=0.5*950*12.5^2/2.2^2
=15334.45 N/m
=15.33 kN/m