Question

A 10.2 gram mass is attached to an ideal spring with a spring constant 30 N/m. The mass is displaced 5.55cm from equilibrium position. What is the maximum velocity of the mass in m/s? Answer to the nearest hundredth. For example 4.26784 m/s would be 4.27m/s

Answer 1

Solution :

Given :

m = 10.2 g = 0.0102 kg

k = 30 N/m

A = 5.55 cm = 0.0555 m

.

According to the principle of conservation of energy : KE = PE

(1/2) m v² = (1/2) k A²

  m v² = k A²

  (0.0102 kg) v² = (30 N/m)(0.0555 m)²

v² = 9.0596 (m/s)²

v = 3.01 m/s