Starting from rest, a 6.70 kg object falls through some liquid and experiences a resistive (drag)
force that is linearly proportional to the velocity of the object. It's measured that the object reaches
half its terminal speed at 4.50 s. a) What is the terminal speed for this object in this liquid? b) At
what time will the speed be 3/4 of the terminal speed? c) How far will the object have traveled in the
first 4.50 s of falling?
(a)
In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving. Therefore,
m•g=r•V, where r is drag coefficient,and V is the terminal force.
V= m•g/r.
Now for accelerated motion,
m•a=m•g-r•v,
dv/dt =g –r•v/m,
Δv/Δt= g –r•v/m,
Since Δv =v(fin)-v(initial), and
v(initial) = 0,
v(fin)=V/2 =>
Δv = V/2.
Then
V/2•Δt= g – r•V/2•m,
m•g/2•r• Δt = g - r•m•g/m•r•2
r=m/ Δt =6.7/4.5 = 1.49 kg/s.
V=mg/r= 67•9.8•4.5/6.7=44.1 m/s.
(b) v=3•V/4.
Δv/Δt= g –r•v/m,
3V/4•Δt =g - r•3•V/4•m,
Solve for Δt,
Δt = 3m/r=3•6.7/1.49 =13.49 s.
(c)
m•a = m•g - m•v,
a = g-r•v/m,
Multiply this equation by Δt.
a• Δt =g• Δt - r•v• Δt /m,
Since a• Δt =Δv=V/2 =22.05 m/s, and
v• Δt = Δs, we obtain
Δs =(m/r)(g• Δt – Δv)=
=(6.7/1.49)(9.8•4.5-22.05)=99 m.
To solve this problem, we'll use the concept of terminal speed and the relationship between drag force and velocity.
Let's start with part a) to find the terminal speed of the object.
Terminal speed occurs when the drag force acting on the object is equal in magnitude and opposite in direction to the weight of the object. At this point, the net force on the object becomes zero, resulting in constant velocity.
When the object reaches half its terminal speed, the net force acting on it is equal to half the weight of the object.
We can write the equation for net force as:
Net force = mg - Fd
where m is the mass of the object (6.70 kg), g is the acceleration due to gravity (9.8 m/s^2), and Fd is the drag force.
When the object reaches half its terminal speed, the drag force Fd is equal to half the weight of the object:
mg/2 = Fd
Now we can write the equation for drag force as:
Fd = kv
where k is a constant of proportionality and v is the velocity of the object.
Substituting the expression for Fd into the equation, we get:
mg/2 = kv
Simplifying the equation, we find:
v = (mg/2k) Equation 1
We can solve for k by using the given information that the object reaches half its terminal speed at 4.50 s.
When t = 4.50 s, the velocity v is half the terminal velocity.
v = (1/2)vt
Now, let's find vt using Equation 1:
(1/2)vt = (mg/2k)
Simplifying the equation further, we get:
vt = (mg/k) Equation 2
From Equation 2, we can see that vt only depends on mass and the constant k. Therefore, vt is the same for all objects falling in the same liquid, regardless of their initial velocities.
Now, we can find the terminal speed (vt) for this object using the given values:
m = 6.70 kg
g = 9.8 m/s^2
Substituting these values into Equation 2, we have:
vt = (6.70 kg * 9.8 m/s^2) / k
Note that without the value of k, we cannot calculate the exact value of terminal speed (vt).
Moving on, let's solve part b) to find the time at which the speed will be 3/4 of the terminal speed.
We have the equation for velocity (v) as:
v = (mg/k) [1 - e^(-kt/m)]
To find the time at which the speed is 3/4 of the terminal speed, we'll substitute the value (3/4)vt for v in Equation 1 and solve for t:
(3/4)vt = (mg/k) [1 - e^(-kt/m)]
Next, we'll rearrange the equation to isolate t:
e^(-kt/m) = 1 - (3/4)
e^(-kt/m) = 1/4
Taking the natural logarithm (ln) of both sides, we have:
-k(4.5 s) / m = ln(1/4)
Simplifying the equation:
-k(4.5 s) / m = ln(1) - ln(4)
-k(4.5 s) / m = -ln(4)
Now, we can solve for t by combining the terms:
t = - (m / 4.5 s) * ln(4/k)
Finally, let's solve part c) to find how far the object will have traveled in the first 4.50 s of falling.
The distance traveled can be found using the formula:
d = (1/2)g * t^2
Substituting the given values, we have:
d = (1/2) * (9.8 m/s^2) * (4.50 s)^2
Simplifying the equation:
d = 98 m
Therefore, the object will have traveled 98 meters in the first 4.50 seconds of falling.