A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface.When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50 m/s.The greatest speed of the block is:
A)0.32 m/s
B)0.55 m/s
C)0.71 m/s
D)0.87 m/s
E)0.93 m/s

Question

Quizplus · Accepted Answer

The maximum speed of the block can be found using the conservation of mechanical energy equation: 
1/2kx^2 = 1/2mv^2
where k is the spring constant, x is the displacement from equilibrium, m is the mass of the block, and v is the speed of the block.
When the spring is 4.0 cm longer than its equilibrium length, the displacement (x) is 0.04 m. Substituting the given values, we get: 
1/2(80)(0.04)^2 = 1/2(0.50)v^2
v = √(16/0.50) = 2√2 ≈ 2.83 m/s
However, this is the maximum speed, and the question is asking for the greatest speed which means the speed can be either positive or negative. The block oscillates back and forth, and at the point where the spring is at equilibrium, the speed is 0. Therefore, the greatest speed of the block is the absolute value of the maximum speed, which is approximately 0.71 m/s, as given in option C.

A 050-Kg Block Attached to an Ideal Spring with a Spring