A 30.0-kg object has a drag force with a magnitude proportional to the square of its speed. The object falls with an acceleration 4.00 m/s2 downward when it is falling downward at 70.0 m/s. What is its terminal velocity?
Answers
At terminal velocity \(v_t\), the drag force has the same magnitude as the object's weight since the object is in equilibrium:
\(F_{\rm net} = F_{\rm drag} - mg = 0 \implies F_{\rm drag} = (30.0\,\mathrm{kg})\left(9.80\dfrac{\rm m}{\mathrm s^2}\right) = 294\,\mathrm N\)
Drag is proportional to the square of the speed, so there is some constant k such that
\(F_{\rm drag} = kv^2\)
and in particular, at terminal velocity we have
\(294\,\mathrm N = k{v_t}^2\)
At the moment the object is falling with speed v = 70.0 m/s and with acceleration a = 4.00 m/s², the net force on it is
\(F_{\rm net}' = F_{\rm drag}' - mg = ma\)
Solve for k :
\(k\left(70.0\dfrac{\rm m}{\mathrm s^2}\right) - (30.0\,\mathrm{kg})\left(9.80\dfrac{\rm m}{\mathrm s^2}\right) = (30.0\,\mathrm{kg})\left(-4.00\dfrac{\rm m}{\mathrm s^2}\right) \\\\ \implies k \approx 0.0355 \dfrac{\rm kg}{\rm m}\)
Now solve for the terminal velocity:
\(294\,\mathrm N = \left(0.0355\dfrac{\rm kg}{\rm m}\right){v_t}^2 \implies \boxed{v_t \approx 91.0\dfrac{\rm m}{\rm s}}\)
Note that the terminal velocity of the object is approximately 600.86 m/s.
How to compute the terminal velocityLet's denote
m = mass of the object = 30.0 kg
a = acceleration of the object = 4.00 m/s² (downward)
v_terminal = terminal velocity (the speed at which the object falls at terminal velocity)
The drag force is given to be proportional to the square of the speed. So, the drag force can be expressed as:
Drag force = k * v²
where k is the constant of proportionality.
At terminal velocity, the drag force and gravitational force are equal in magnitude but opposite in direction:
Drag force = Gravitational force
k * v_terminal^2 = m * g
where g is the acceleration due to gravity (approximately 9.81 m/s²).
Now, we can solve for v_terminal:
v_terminal^2 = (m * g) / k
v_terminal = √((m * g) / k)
Given that v = 70.0 m/s when the object is falling downward at 70.0 m/s:
70.0 m/s = v_terminal
Now, we can solve for k using the initial acceleration:
4.00 m/s² = k * v^2
k = 4.00 m/s² / v^2
Substitute the value of v into the equation for k:
k = 4.00 m/s² / (70.0 m/s)^2
Now, calculate k:
k = 4.00 m/s² / 4900 m²/s² = 8.16 x 10^(-4) s²/m²
Finally, calculate the terminal velocity:
v_terminal = √((m * g) / k)
= √((30.0 kg * 9.81 m/s²) / (8.16 x 10^(-4) s²/m²))
= √(294.3 m²/s² / 8.16 x 10^(-4) s²/m²)
= √(361083.333 s²)
≈ 600.86 m/s
The terminal velocity of the object is approximately 600.86 m/s.
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