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When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.
a. what is the force constant of the spring.
b. if the 2.50?

Respuesta :

austinbaird
F=K*X,
F=M*a 

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant 

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m 


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