Two springs of force constants 300 ~N / m (Spring A) and 400 ~N / m (Spring B) are joined together in series. The combination is compressed by 8.75 ~cm . The ratio of energy stored in A and B is E_A E_B . Then E_A E_B is equal to:

Step-by-step solution

Given : k _ A =300 ~N / m , k _ B =400 ~N / m Let when the combination of springs is compressed by force F. Spring A is compressed by x . Therefore compression in spring B aligned & x_B=(8.75-x) cm & F =300 x =400(8.75- x ) aligned Solving we get, x=5 ~cm x _ B =8.75-5=3.75 ~cm E _ A E _ B = 1 2 k _ A ( x _ A )^2 1 2 k _ B ( x _ B )^2 = 300 (5)^2 400 (3.75)^2 = 4 3