A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200 N/m.When the spring has its equilibrium length the block is given a speed of 5 m/s.What is the maximum elongation of the spring?

Question

Quizplus · Accepted Answer

The maximum elongation of the spring can be found using the conservation of energy principle. Initially, the block has kinetic energy only, which is given by:
KE = (1/2)mv^2 = (1/2)(2 kg)(5 m/s)^2 = 25 J
As the block comes to rest at the maximum elongation of the spring, all the initial kinetic energy is converted to potential energy stored in the spring, which is given by:
PE = (1/2)kx^2
where k is the spring constant and x is the elongation of the spring from its equilibrium length. Therefore, we have:
25 J = (1/2)(200 N/m)x^2
x^2 = 0.25 m^2
x = 0.5 m (since x is positive, the negative root is discarded)

A 2-Kg Block Is Attached to a Horizontal Ideal Spring