A system consists of two springs connected in series and each having the spring constant 10 Nm ⁻¹ . The minimum work required to stretch this system by l cm in erg is

Step-by-step solution

For series combination, effective force constant is given by 1 K_ eff = 1 K₁ + 1 K₂ Here, K₁=K₂=10 ~N / m 1 K_ eff = 1 10 + 1 10 = 1 5 or K_ eff =5 ~N / m = 5 10^5 100 dyne / cm K_ eff =5 10^3 dyne / cm Work done in stretching this spring, W= 1 2 K_ eff x^2= 1 2 5 10^3 1^2=2500 erg