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