An
object of mass m is launched from a stationary helicopter towards the Earth with
the speed v_{0}. It experiences a force of air resistance F = -kv,
where k is a positive constant. The positive direction of all vector
quantities is downward.

(a) Draw a free body diagram showing the forces acting on the object.

(b) What is the terminal speed of the object?

(c) Find the speed of the object as a function of time, v_{0}, m, g and
k.

(d) What is the direction of the acceleration of the object?

Solution:

- Concepts:

Newtonian mechanics - Reasoning:

**F**_{net }= m**a**is the equation of motion for the object - Details of the calculation:

(a) Since velocity is downward, air resistance is upward, in the opposite direction of gravity.

(b) When the object reaches terminal velocity, its acceleration is a = 0 and the sum of all forces acting on the object is 0.

0 = mg - kv_{t}, v_{t}= mg/k.

(c) a = dv/dt

m(dv/dt) = mg – kv, dv/(v – mg/k) = -(k/m)dt

∫ dv/(v – mg/k) = -∫(k/m)dt

ln(v - mg/k) - ln(v_{0}- mg/k) = -(k/m)t

v = mg/k + (v_{0}- mg/k)e^{-(k/m)t}

(d) a = dv/dt

a = (-k/m)(v_{0}- mg/k)e^{-(k/m)t}

a = (g - v_{0}k/m)e^{-(k/m)t}

The acceleration can be positive, negative or zero.

If v_{0}< v_{t}= mg/k, the acceleration is positive or downward

If v_{0}> v_{t}= mg/k, the acceleration is negative or upward.

If v_{0}= v_{t}= mg/k, the acceleration is zero.

A particle of mass m falls subject to the pull of gravity, with acceleration
g, and to the force of air resistance. The particle is dropped from height z =
z_{0}. The initial velocity is zero. The force of air resistance may be modeled
as linearly dependent on the speed, so that the height of the projectile
satisfies

md^{2}z/dt^{2} = -mg - bdz/dt.

(a) Solve the equation of motion for the particle's height z(t). Find the
terminal velocity.

(b) Take the b --> 0 limit and show that usual free-fall solution is obtained.

Solution:

- Concepts:

Newton's second law, integrating the equation of motion - Reasoning:

The equation of motion is given. - Details of the calculation:

(a) Let v = dz/dt, then mdv/dt = -mg – kv. Try a solution v = A + Be^{-kt}.

v(t) = -(g/k) + (g/k)e^{-kt}, where k = b/m.

As t approaches infinity, v approaches v_{termial}= -g/k.

z(t) = z_{0}+ ∫_{0}^{t}(-(g/k) + (g/k)e^{-kt'})dt' = z_{0}- (g/k)t + (g/k^{2})(1 - e^{-kt}).

(b) As b --> 0, k --> 0, e^{-kt}= 1 – kt + k^{2}t^{2}/2 +… .

z(t) = z_{0}- (g/k)t + (g/k^{2})( kt - k^{2}t^{2}/2) = z_{0}– gt^{2}/2.

The forces acting on a sky-diver of mass m are the force of gravity and the
drag force due to the air. Assume the drag force is proportional to the
square of the speed. Find the diver's velocity as a function of time, and
the diver's terminal velocity v_{f}. Assume v_{i} = 0.

Hint: ∫dx/(a^{2} – x'^{2}) = (1/a) tanh^{-1}(x/a)

Solution:

- Concepts:

Newton's second law - Reasoning:

Let the diver's mass be m and the coefficient of proportionality between the drag force and v^{2}be mk.

mdv/dt = mg – mkv^{2}. - Details of the calculation:

Terminal velocity v_{f}= V: dv/dt = 0 --> V^{2}= g/k.

In terms of V, dv/dt = g(1 – v^{2}/V^{2}), dv/(V^{2}– v^{2}) = (g/V^{2})dt.

∫_{0}^{v}dv'/(V^{2}– v'^{2}) = gt/V^{2}.

Integrating and solving for v we get

v(t) = V tanh(gt/V).

A boat with mass m is slowed by a drag force F(v). Its velocity decreases
according to the formula v(t) = c^{2}(t – t_{f})^{2} for
t ≤ t_{f}, where c is a constant and t_{f} is the time a which
it stops. Find the force F(v) as function of v.

Solution:

- Concepts:

Newton's second law - Reasoning:

F = mdv/dt - Details of the calculation:

a(t) = dv/dt = 2c^{2}(t – t_{f})

F(t) = = ma(t) = 2mc^{2}(t – t_{f})

t – t_{f}= -√(v)/c (Select the negative solution since t – t_{f }is negative.)

F(v) = -2mc^{2}√(v)/c = -2mc√(v)