Questions: An object of mass 10 kg is released from rest 2000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b=50 N sec/m, determine the equation of motion of the object. When will the object strike the ground? Assume that the acceleration due to gravity is 9.81 m / sec^2 and let x(t) represent the distance the object has fallen in t seconds. Determine the equation of motion of the object. x(t)=

Solution Steps

The forces acting on the object are gravity and air resistance. The gravitational force is given by $F_g = mg$, where $m = 10 \, \text{kg}$ and $g = 9.81 \, \text{m/s}^2$. The force due to air resistance is $F_r = -bv$, where $b = 50 \, \text{N s/m}$ and $v$ is the velocity.

The net force $F$ on the object is: $$F = mg - bv$$

According to Newton's second law, $F = ma$, where $a = \frac{dv}{dt}$. Therefore, we have: $$m \frac{dv}{dt} = mg - bv$$

Substituting the given values: $$10 \frac{dv}{dt} = 10 \times 9.81 - 50v$$

Simplifying, we get the differential equation: $$\frac{dv}{dt} = 9.81 - 5v$$

This is a first-order linear differential equation. We can solve it using separation of variables: $$\frac{dv}{9.81 - 5v} = dt$$

Integrating both sides: $$\int \frac{1}{9.81 - 5v} \, dv = \int \, dt$$

The left side integrates to: $$-\frac{1}{5} \ln |9.81 - 5v| = t + C$$

Solving for $v$, we exponentiate both sides: $$|9.81 - 5v| = e^{-5t + C}$$

Let $C_1 = e^C$, then: $$9.81 - 5v = C_1 e^{-5t}$$

Solving for $v$: $$v = \frac{9.81}{5} - \frac{C_1}{5} e^{-5t}$$

Since the object is released from rest, $v(0) = 0$. Substituting into the velocity equation: $$0 = \frac{9.81}{5} - \frac{C_1}{5}$$

Solving for $C_1$: $$C_1 = 9.81$$

Thus, the velocity equation becomes: $$v(t) = \frac{9.81}{5} (1 - e^{-5t})$$

The position function $x(t)$ is the integral of the velocity function: $$x(t) = \int v(t) \, dt = \int \frac{9.81}{5} (1 - e^{-5t}) \, dt$$

Integrating: $$x(t) = \frac{9.81}{5} \left( t + \frac{1}{5} e^{-5t} \right) + C_2$$

Since the object starts from rest at $x(0) = 0$: $$0 = \frac{9.81}{5} \left( 0 + \frac{1}{5} \right) + C_2$$

Solving for $C_2$: $$C_2 = -\frac{9.81}{25}$$

Thus, the position function is: $$x(t) = \frac{9.81}{5} \left( t + \frac{1}{5} e^{-5t} \right) - \frac{9.81}{25}$$

Final Answer