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o-na [289]
3 years ago
14

A 10-kg mass, hung by an ideal spring, causes the spring to stretch 2.0 cm. what is the spring constant (force constant) for thi

s spring?
Physics
2 answers:
Lorico [155]3 years ago
8 0
From Hooke's law the spring constant k is given by F = ke; then k = F/e. But F = Ma = Mg = 10 * 9.8 = 98N where g = 9.8ms^(-2) Extension e = 2cm = 0.02m. So k = 98/0.02 = 4900 N/m.
Veseljchak [2.6K]3 years ago
8 0
The computation for this problem is calculated by:F = kx 
F = (10)(9.8) 
x= .02 meters after converting from cm to m 
98 = (k) (.02) So, k=4900 N m.
Then back to cm by dividing by 100 which is now 49 N cm would be the spring constant for this problem .
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