Question

an object of mass 0.2 kg is hung from a spring whose spring constant is 80 n/m. the object is subject to a resistive force given by −, where is its velocity in meters per second.

Answer 1

The object's velocity as a function of time is given by the equation
`\(v(t) = (9.8)/(8) \left(1 - e^{-(8)/(5)t}\right)\)`meters per second.

Step-by-step explanation:

The equation describing the motion of the object is derived from the differential equation
`\(m (dv)/(dt) + c v + kx = 0\),` where (m) is the mass of the object, (c) is the coefficient of resistive force, (k) is the spring constant, (x) is the displacement from the equilibrium position, and (v) is the velocity.

Given that the mass (m = 0.2 kg), the spring constant (k = 80 N/m), and the resistive force (c = 4 Ns/m), the differential equation becomes
`\(0.2 (dv)/(dt) + 4v + 80x = 0\).` Since the object is hung vertically, (x) is the displacement from the equilibrium position, and (x) can be expressed as
`\(x(t) = -(1)/(8) \left(1 - e^{-(8)/(5)t}\right)\)`. Substituting this into the equation and solving for (v(t)), we get the final answer.

The initial conditions of the system, such as the initial displacement and velocity, are not provided. Therefore, the solution is presented in terms of time (t), and additional information would be needed to determine specific values. The solution represents the velocity of the object at any given time (t) in the context of the provided parameters.