Question

A 1.74-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 n/m. the object is displaced 3.50 m to the right from its equilibrium position and then released, initiating simple harmonic motion.

(a) what is the force (magnitude and direction) acting on the object 3.50 s after it is released? magnitude n direction direction

Answer 1

To find the force acting on a 1.74-kg object undergoing SHM 3.50 s after release, equations involving the object's displacement, spring constant, and mass are used to determine the position function x(t) and subsequently the force F = -kx.

Step-by-step explanation:

The question is related to an object undergoing simple harmonic motion (SHM) on a frictionless horizontal track, due to a spring with a known force constant.

To solve the mathematical problem completely and find the force acting on the object 3.50 seconds after it is released, we use the equation for the force on a spring system in SHM, F = -kx, where k is the spring constant and x is the displacement from equilibrium.

However, we must first determine the displacement at 3.50 seconds using the SHM position function x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. With the given force constant (k = 5.00 N/m) and the mass of the object (1.74 kg), we can calculate ω using the formula ω = √(k/m).

Then, knowing the object was released from rest at the maximum amplitude, φ would be zero, and the displacement at the given time can be calculated. Using this displacement, we can then calculate the force acting on the object.