A 2.24 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.45 N/m. The mass is displaced 3.06 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the force (including direction, choose the right to be positive) acting on the mass 4.20 s after it is released
To find the force acting on the mass 4.20 seconds after it is released, we need to determine the position of the mass at that time.
In simple harmonic motion, the position of the mass as a function of time can be described by the equation x(t) = A * cos(ωt + φ), where x(t) is the position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.
First, let's determine the angular frequency. The formula for angular frequency is ω = √(k/m), where k is the force constant of the spring and m is the mass. Plugging in the values given:
ω = √(4.45 N/m / 2.24 kg) ≈ 2.49 rad/s
Next, let's determine the amplitude. The amplitude (A) is the maximum displacement from the equilibrium position. In this case, the mass is displaced 3.06 m to the right, so the amplitude is:
A = 3.06 m
Now, let's find the phase constant φ. The phase constant determines the initial position and velocity of the mass. Since the mass is initially displaced 3.06 m to the right, the phase constant φ is zero.
Now, we can use the position equation to find the position of the mass at t = 4.20 s:
x(4.20 s) = A * cos(ω * t + φ)
= 3.06 m * cos(2.49 rad/s * 4.20 s)
Evaluating this expression:
x(4.20 s) ≈ 3.06 m * cos(10.47 rad)
≈ 3.06 m * (-0.907)
Therefore, the position of the mass at 4.20 seconds is approximately -2.78 m.
Now, we can calculate the force acting on the mass using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from equilibrium.
F = -k * x
Plugging in the values:
F = -4.45 N/m * (-2.78 m)
≈ 12.39 N
Since the mass is released to the right, the force acting on the mass 4.20 seconds after it is released is approximately +12.39 N to the right.