A block is placed in oscillation on a frictionless horizontal surface attached to a spring that requires 125 N for the block to be stretched 15 cm away from equilibrium.From this position it is released,and oscillates with a period of 0.25 s.What is the mass of the block?
A) 0.66 kg
B) 4.1 kg
C) 1.3 kg
D) 5.3 kg

Question

Quizplus · Accepted Answer

We can use the formula for the period of oscillation of a spring-mass system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

We are given that the period is 0.25 s, so we can rearrange the equation to solve for the mass:

m = (T²k)/(4π²)

To find k, we need to convert the given stretch distance into a force. The spring constant k is defined as the force required to stretch the spring by one unit of distance. So if the block needs a force of 125 N to be stretched 15 cm, we can divide 125 by 0.15 to get:

k = 833.3 N/m

Now we can plug in the values for T and k and solve for m:

m = (0.25² x 833.3)/(4π²) = 1.3 kg

Therefore, the mass of the block is 1.3 kg.

A Block Is Placed in Oscillation on a Frictionless Horizontal