A 1200 kg car rolling on a horizontal suraface has a speed u= 65km/h when it strikes a horizontal coiled spring and is bropught to rest in a distance of 2.2 m. what is the spring stiffnes constant of the spring?
Find the kinetic energy (KE = 1/2 mv^2) the car has just before it hits the spring. That kinetic energy will all be converted into elastic potential energy (PE = 1/2 kx^2), when the spring has been compressed 2.2m. Convert km/h to m/s first, (divide km/h by 3.6).
18.0555555556 m/s ,then what's next?
The steps are above :P
The KE will all become PE... so: 1/2 mv^2 = 1/2 kx^2
how to get the x?
The x is the distance the spring is stretched or compressed.. you're given x.
ahhh.. how about k?
Do algebra to find k. It's the spring constant.
ahh..okay...
Or just plug in the numbers you do have, and do algebra to find k.
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