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Mathematics 17 Online
OpenStudy (anonymous):

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?

OpenStudy (agent0smith):

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).

OpenStudy (anonymous):

18.0555555556 m/s ,then what's next?

OpenStudy (agent0smith):

The steps are above :P

OpenStudy (agent0smith):

The KE will all become PE... so: 1/2 mv^2 = 1/2 kx^2

OpenStudy (anonymous):

how to get the x?

OpenStudy (agent0smith):

The x is the distance the spring is stretched or compressed.. you're given x.

OpenStudy (anonymous):

ahhh.. how about k?

OpenStudy (agent0smith):

Do algebra to find k. It's the spring constant.

OpenStudy (anonymous):

ahh..okay...

OpenStudy (agent0smith):

Or just plug in the numbers you do have, and do algebra to find k.

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