Question

Two identical springs have the same force constant of 147 Newton meter. What elongation will be produced in each case ?

Answer 1

Explanation:

Given Two identical springs have the same force constant of 147 Newton meter. What elongation will be produced in each case ?

• We know that F = - kx (x is the displacement)
• So y = F / k (effective field)
• Now two springs connected in parallel.
• Effective k = k1 + k2
•                 = 147 + 147 N/m
•               = 294 N/m
• Now effective y = mg
•                            = 5 x 9.8 / 294 m (mass = 5 kg)
•                           = 49 / 294
•               Or y = 1/6 m
• Springs are connected in series,
• So 1/ k effective = 1/k1 + 1/k2
• Or k effective = k1 k2 / k1 + k2
•                       = 147 x 147 / 2 x 147
•                       = 147 / 2 N/m
• So y = 5 x 9.8 x 2 / 147
•        = 49 x 2 / 147
• Or y = 2/3 m  
• So y = 5 x 9.8 / 147
•      So y = 49 / 147
• Or y = 1/3 m

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Answer 2

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