identical springs are placed side-by-side (in parallel), and connected to a large massive block. The stiffness of the 43-spring
1 answer:
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 = 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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<h2>Answer:</h2>
4E
<h2>Explanation:</h2>The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by
U = --------------------(i)
Where;
U = potential energy stored
k = spring constant of the material
x = elongation (extension or compression of the material).
<em>From the first statement;</em>
<em>when elongation (x) is 4cm, energy stored (U) is E</em>
<em>Substitute these values into equation (i) as follows;</em>
E =
E = 8k
<em>Make k subject of the formula</em>
k = [measured in J/cm]
<em>From the second statement;</em>
<em>It is stretched by 4cm.</em>
This means that total elongation will be 4cm + 4cm = 8cm.
The potential energy stored will be found by substituting the value of x = 8cm and k = into equation (i) as follows;
U =
U =
U =
Therefore, the potential energy stored will now be 4 times the original one.