A 9.00 kg object starting from rest falls through a viscous medium and experiences a resistive force = -b , where is the velocity of the object. The object reaches one half its terminal speed in 5.54 s.
(a) Determine the terminal speed.
(b) At what time is the speed of the object three-fourths the terminal speed?
(c) How far has the object traveled in the first 5.54 s of motion?
PLEACE, HELP ME
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To solve this problem, we can use the equation for the resistive force experienced by the object:
F = -bv
where F is the resistive force, b is the constant, and v is the velocity of the object.
(a) To find the terminal speed, we can use the fact that the object reaches one half its terminal speed in 5.54 s. At terminal speed, the resistive force is equal in magnitude and opposite in direction to the gravitational force acting on the object.
Therefore, we can equate the gravitational force and the resistive force at terminal velocity:
mg = bv_terminal
where m is the mass of the object and g is the acceleration due to gravity. Rearranging the equation gives us:
v_terminal = mg / b
Substituting the given values:
m = 9.00 kg
g = 9.81 m/s^2
We need to know the value of b in order to calculate the terminal speed.
(b) To find the time at which the speed is three-fourths the terminal speed, we can use the fact that the object reached one half its terminal speed in 5.54 s.
We know that at time t = 5.54 s, the speed of the object is one-half the terminal speed. Let's denote this speed as v_half.
So, v_half = (1/2) * v_terminal
We want to find the time when the speed is three-fourths the terminal speed. Let's denote this speed as v_three_fourths.
So, v_three_fourths = (3/4) * v_terminal
We need to find the time t_three_fourths when the speed is three-fourths the terminal speed.
(c) To find how far the object has traveled in the first 5.54 s of motion, we can use the equation for distance traveled during constant acceleration:
d = (1/2) * a * t^2
where d is the distance traveled, a is the acceleration, and t is the time.
We need to find the values of the acceleration and time to calculate the distance traveled.
Let's go step-by-step to solve each part of the problem.
(a) Determine the terminal speed:
v_terminal = mg / b
v_terminal = (9.00 kg) * (9.81 m/s^2) / b
(b) At what time is the speed of the object three-fourths the terminal speed:
We need to find the value of t_three_fourths when the speed is three-fourths the terminal speed.
(c) How far has the object traveled in the first 5.54 s of motion:
d = (1/2) * a * t^2
We need to find the values of the acceleration and time to calculate the distance traveled.
To determine the answers to these questions, we need to use the equations of motion for an object experiencing a resistive force. Here's how we can approach each part:
(a) To find the terminal speed, we can use the fact that the object reaches half its terminal speed in 5.54 s. At half the terminal speed, the net force on the object is zero. The net force can be calculated by subtracting the resistive force from the gravitational force.
The gravitational force is given by:
F_grav = m * g
where m is the mass of the object (9.00 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
The resistive force is given by:
F_resist = -b * v
where b is the resistance constant and v is the velocity of the object.
Setting the net force to zero, we have:
F_grav + F_resist = 0
m * g - b * v = 0
Since we are looking for the terminal speed, we know that at terminal velocity, the net force is zero. Therefore, the gravitational force equals the resistive force:
m * g = b * v_terminal
Solving for v_terminal, we get:
v_terminal = (m * g) / b
Substituting the values, we have:
v_terminal = (9.00 kg * 9.8 m/s²) / b
(b) To find the time at which the speed of the object is three-fourths the terminal speed, we can set up a ratio using the information given. We know that the object reaches half its terminal speed in 5.54 s. So, the time it takes to reach three-fourths the terminal speed will be 3/2 times the time it took to reach half the terminal speed.
Let's call the time at which the speed is three-fourths the terminal speed as t_3/4th. We can set up the following equation:
t_3/4th = (3/2) * 5.54 s
(c) To find how far the object has traveled in the first 5.54 seconds of motion, we can use the equation of motion for uniformly accelerated motion:
d = (1/2) * a * t²
where d is the distance traveled, a is the acceleration, and t is the time.
In this case, the acceleration is the net force divided by the mass of the object:
a = (F_grav + F_resist) / m
We can substitute the force equations and simplify to get:
a = (m * g - b * v) / m
Now, we can use this acceleration value in the equation for distance to find how far the object has traveled in the first 5.54 seconds.
d = (1/2) * a * (5.54 s)²
Now, you have the step-by-step process to find the answers to all the parts of the question. Just plug in the given values and solve the equations to find the terminal speed, the time at three-fourths the terminal speed, and the distance traveled in the first 5.54 seconds.