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Hi. I just solved a homework problem. I eventually got the right answer and am only confused about why it was solved that way. This is the reason I posted it here instead of in the homework help section. All I need is a little insight into why the solution is the way it is. OK here we go:
A 1kg block sits atop a 5kg block as it osscilates on a spring. The period is 1.5s. The small block begins to slip when the Amplitude is increased to .4m What is the coefficient of satic friction bewteen the two blocks.
The solution is simple:
[tex] F_{sp} = f_s [/tex]
[tex] -kA = \mu_smg [/tex]
[tex] \mu_s = \frac{-kA}{mg} [/tex]
Using the period you can solve for k = 105.3 N/m. The you can solve for [tex] \mu_s [/tex]
Now here's my problem. In order to get the right answer you must use m= the total mass of the system= 6kg. I thought that since only the small block was sliding, only its mass shouldbe used in this calculation. Why is the total mass used? That is my question. Thank you for your help. (If this should still be in the homework help section, please move it.)
A 1kg block sits atop a 5kg block as it osscilates on a spring. The period is 1.5s. The small block begins to slip when the Amplitude is increased to .4m What is the coefficient of satic friction bewteen the two blocks.
The solution is simple:
[tex] F_{sp} = f_s [/tex]
[tex] -kA = \mu_smg [/tex]
[tex] \mu_s = \frac{-kA}{mg} [/tex]
Using the period you can solve for k = 105.3 N/m. The you can solve for [tex] \mu_s [/tex]
Now here's my problem. In order to get the right answer you must use m= the total mass of the system= 6kg. I thought that since only the small block was sliding, only its mass shouldbe used in this calculation. Why is the total mass used? That is my question. Thank you for your help. (If this should still be in the homework help section, please move it.)