Consider the two ideal springs connected end-to-end and then to a box as shown a

Question

Consider the two ideal springs connected end-to-end and then to a box as shown above. The two springs have spring constants K1 = 30 N/m and K2 = 40 N/m respectively. If the inertia of the box is 7 kg, what will be the total stretch of the springs when the box is hung from them? (In other words, if the unstretched length of the two springs combined is L and the length of the two springs with the box hanging is L + d, what is d?) Hint: What are the tensions in the two springs? Give your answer in meters to three significant figures.

Explanation / Answer

Weight W = mg = 7*9.81 = 68.67 N

F = k1*x1 = k2*x2

68.67 = 30*x1 = 40*x2

So, x1 = 68.67/30 = 2.289 m

x2 = 68.67/40 = 1.71675 m

Total stretch = x1 + x2 = 2.289 + 1.71675 = 4.00575 m

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