When stretched by 25 cm , a spring stores 5 J of energy. It is positioned vertically with its lower end fixed. A block secured to the opposite end is made to undergo small oscillations. If the block completes 5 oscillations each second, determine the mass of the block. (Assume ^2 10 )

Step-by-step solution

The potential energy stored in the spring is given by U = 1 2 kx^2 . Substituting the given values, U = 5 J and x = 25 cm = 0.25 m : 5 = 1 2 k (0.25)^2 5 = 1 2 k ( 1 16 ) k = 5 32 = 160 N/m The frequency of oscillation is f = 5 Hz . The angular frequency is = 2 f = 10 rad/s . Using the formula for angular frequency = k m : ^2 = k m (10 )^2 = 160 m 100 ^2 = 160 m Given ^2 10 : 1000 = 160 m m = 160 1000 = 0.16 kg