A body of mass m falling from rest is subjected to the force of gravity and an air resistance proportional to the square of the velocity. If it falls through a distance x and posseses a velocity v at that instant then show that.
To solve this problem, we need to analyze the forces acting on the falling body, including gravity and air resistance.
The force of gravity can be given by:
F_gravity = mg
where m is the mass of the body and g is the acceleration due to gravity.
The force of air resistance is proportional to the square of the velocity, so it can be expressed as:
F_air = -kv^2
where k is the constant of proportionality.
Using Newton's second law, the net force can be written as:
F_net = ma
where a is the acceleration of the body.
Since the body is falling vertically, we can write the acceleration as the second derivative of the position with respect to time:
a = d^2x/dt^2
Now, we can write the equation of motion as:
F_net = ma = mg - kv^2
Since the velocity v is given as the final velocity at a particular position x, we can write:
v = dx/dt
Taking the first derivative of the equation of motion with respect to time, we get:
dv/dt = d^2x/dt^2
Substituting these expressions, we have:
ma = mg - kv^2
m(dv/dt) = mg - k(dx/dt)^2
Now, let's solve this equation step by step.
1. Start by isolating the terms involving velocity on one side:
m(dv/dt) + kv^2 = mg
2. Divide both sides by m:
dv/dt + (k/m)v^2 = g
3. Separate the variables by multiplying both sides by dt:
1/(g dt) dv/(1 + (k/m)v^2) = 1
4. Integrate both sides with respect to v:
∫1/(1 + (k/m)v^2) dv = ∫1/(g dt)
5. The integral on the left-hand side can be solved using partial fraction decomposition or by using the substitution u = v √(k/m):
∫[1/(1 + u^2)] (du/√(k/m)) = ∫1/(g dt)
6. The integral on the left-hand side is the inverse tangent function:
arctan(u √(k/m)) = (1/g) t + C
where C is the constant of integration.
7. Substituting back for u:
arctan(v √(k/m)) = (1/g) t + C
8. To solve for v, take the tangent of both sides:
v √(k/m) = tan((1/g) t + C)
9. Finally, solve for v:
v = √(m/k) tan((1/g) t + C)
This is the expression for the velocity v of the falling body at any time t. The constant C can be determined using the initial conditions of the problem, such as the initial velocity and position.
To show the relationship between the distance fallen, velocity, and mass of the body, we will use Newton's Second Law of Motion and the principles of kinematics.
1. Newton's Second Law of Motion states that the net force acting on an object is equal to the product of its mass (m) and its acceleration (a). Mathematically, this can be expressed as:
F_net = m * a
2. In this scenario, there are two forces acting on the falling body: gravity (mg, where g is the acceleration due to gravity) and air resistance (kv^2, where k is the proportionality constant and v is the velocity of the body). Therefore, the net force can be written as:
F_net = mg - kv^2
3. According to kinematics, the acceleration (a) of an object can be calculated as the derivative of its velocity (v) with respect to time (t). Since we are interested in the relationship between distance (x) and velocity (v), we need to express acceleration in terms of these variables.
a = dv/dt
4. The definition of velocity states that velocity (v) is the derivative of distance (x) with respect to time (t). Hence, we can express acceleration in terms of x and v by differentiating velocity with respect to time:
a = d^2x/dt^2
5. Now, equating the expressions for acceleration from step 3 and step 4, we have:
d^2x/dt^2 = dv/dt
6. Rearranging equation 5, we get:
dv = d^2x/dt^2 * dt
7. Integrating both sides of equation 6 with respect to time, we obtain:
∫dv = ∫d^2x/dt^2 * dt
This simplifies to:
v = dx/dt
8. Integrating both sides of equation 7 with respect to distance, we get:
∫v dx = ∫dx/dt * dx
The left-hand side of this equation represents the work done W, which is given by the change in potential energy of the body:
W = ∫mg dx - ∫kv^2 dx
Since ∫mg dx is equal to the negative of the change in potential energy (mgh), we have:
W = -mgh - ∫kv^2 dx
Note that the negative sign appears because the work done by gravity decreases the potential energy.
9. Now, let's consider the relationship between work done and kinetic energy (KE). According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. Mathematically, this can be expressed as:
W = ΔKE
10. At the beginning (when the body is at rest), all the potential energy is converted to kinetic energy as it falls. Therefore, W is equal to the change in kinetic energy (ΔKE), which is given by:
W = KE_final - KE_initial
Since the body starts from rest, KE_initial is zero. Hence:
W = KE_final - 0
W = KE_final
11. At the instant when the body has fallen through a distance x and possesses a velocity v, its kinetic energy can be expressed as:
KE_final = (1/2)mv^2
12. Substituting equations 11 and 10 into equation 9, we obtain:
(1/2)mv^2 = -mgh - ∫kv^2 dx
13. Rearranging equation 12, we get:
(1/2)mv^2 + ∫kv^2 dx = -mgh
14. The integral on the left-hand side represents the work done by air resistance during the fall. Therefore, we can write it as:
Work_done_by_air_resistance = ∫kv^2 dx
15. Applying the chain rule, we can change the integration variable from dx to v:
dx = v * dt
Substituting this into equation 14, we get:
Work_done_by_air_resistance = ∫kv^2 * v * dt
16. Simplifying equation 15, we have:
Work_done_by_air_resistance = k * ∫v^3 dt
17. Integrating v^3 with respect to t, we obtain:
∫v^3 dt = (1/4)v^4 + C
where C is the constant of integration.
18. Substituting equation 17 into equation 16, we have:
Work_done_by_air_resistance = k * ((1/4)v^4 + C)
19. Finally, substituting equations 18 and 13 into equation 12, we get:
(1/2)mv^2 + k * ((1/4)v^4 + C) = -mgh
This equation relates the distance fallen (x), velocity (v), and mass (m) of the falling body.
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