A 50 g mass hanger hangs motionless from a partially stretched spring. When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm. What is the spring constant of the spring (in N/m)? (Assume g = 9.79 m/s2.

Question

A 50 g mass hanger hangs motionless from a partially stretched spring.  When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm.  What is the spring constant of the spring (in N/m)?  (Assume g = 9.79 m/s2.

anonymous · Answer

First, find the change in force when the second mass is added-->change in F=(m2+m1)g-m1g

deltaF=115g-50g=65g (convert 65 g to kg) = 0.63635 N

Now, this resulted in the stretching of the spring by 10 cm (0.1 m)-->
deltaF=kx=k*0.1-->k=spring constant=0.157 N/m

anonymous · Answer

Thank you!

anonymous · Answer

you should recheck...
at the first rest position, kx = mg, (where x is the initial stretch when the mass hanger is hung) or else it would not be at rest. For the combination of mass hanger and mass $$k(x + x') = (m + M)g$$where x' is the additional stretch (0.1m). Expand$$kx+kx'=mg+Mg$$but we already know that with the mass hanger alone kx=mg, so$$mg+kx'=mg+Mg$$regrouping$$kx'=Mg$$or$$k=Mg/x'=(9.8m/s^{2})(0.065kg)/(0.1m)=6.37N/m$$If the above answer was true, then a force of 0.157 newtons would stretch the spring 1 meter (0.157N/m), but the problem applies a force of 0.065kg*9.8m/s^2 = 0.637N and it stretches only 0.1m... a k of 0.157N/n would say it should stretch 0.637N/(0.157N/m) = 4m!