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Flauer [41]
3 years ago
7

identical springs are placed side-by-side (in parallel), and connected to a large massive block. The stiffness of the 43-spring

combination is 16770 N/m. What is the stiffness of one of the individual springs? ks= N/m
Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

The stiffness of one of the individual spring is 390 N/m.

Explanation:

It is given that 43 identical springs are placed side-by-side and connected to a large massive block.

The stiffness of the 43 spring combination is 16770 N/m

We need to find the stiffness of one of the individual springs. Let k is the stiffness of one spring. The effective spring stiffness of this width spring is given by :

K=N\times k

k=\dfrac{K}{N}

k=\dfrac{16770\ N/m}{43}

k = 390 N/m

So, the stiffness of one of the individual spring is 390 N/m. Hence, this is the required solution.

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