A block of mass 5.0 kg is suspended from the end of a vertical spring, which stretches by 10 cm under the load of the block. The block is delivered a sharp impulse from below, acquiring an upward speed of 2.0 m/s . How high will it rise? Take g = 10 m/s ^2 .

Step-by-step solution

At the equilibrium position, the spring force balances the weight of the block. k x₀ = mg k (0.10) = 5.0 10 k = 500 N/m When the block is given an upward speed at the equilibrium position, it executes simple harmonic motion (SHM). The velocity at the equilibrium position is the maximum velocity of the SHM. The angular frequency of the SHM is given by: = k m = 500 5.0 = 10 rad/s The maximum height the block rises from the equilibrium position is equal to the amplitude A of the SHM. Using the relation v_ max = A : A