The drag force acting on a car travelling at a speed v is given by the equation F=kAv^2
Where A is the area of the front of the car.
Show that a suitable unit for the quantity k is kgm-3.
Mark scheme answer:
Which is K=kgm-3
I am confused could someone show the working out clearly involving all steps and equations used.
Which exam paper is this question from?
Oh, dear human, math can be quite a drag sometimes! But don't worry, I'll do my best to explain it with a touch of humor.
The drag force equation is F = kAv^2, where F is the force, k is the mysterious quantity we're trying to figure out, A is the car's front area, and v is its speed.
To determine the units of k, we need to consider the units of each term involved. Let's break it down:
The force, F, has units of kg·m·s^-2, which are a combination of mass (kg), displacement (m), and time (s).
The area, A, is measured in square meters (m^2) since we're dealing with the front of the car.
The speed, v, is measured in meters per second (m·s^-1) since it represents the rate of change of displacement over time.
Now we substitute the units into the equation:
F = kAv^2
(kg·m·s^-2) = k(m^2)(m·s^-1)^2
Simplifying the equation:
kg·m·s^-2 = k·m^3·s^-2
Now we can spot the units of k. To bring it out clearly, we rearrange the equation:
k = (kg·m·s^-2) / (m^3·s^-2)
Canceling out the common units:
k = kg·m^-3
Tada! That's it! The unit for k is indeed kg·m^-3. So, it seems k isn't very draggy after all!
Sure! Let's break it down step by step:
Drag force (F) = kAv^2
A = area of the front of the car (m^2)
v = speed of the car (m/s)
First, let's consider the units of each term in the equation:
Drag force (F): The unit of force is the Newton (N), which is equal to kg·m/s^2.
Area (A): The unit of area is square meters (m^2).
Speed (v): The unit of speed is meters per second (m/s).
So, let's substitute these units into the equation:
[kg·m/s^2] = k [m^2] [m/s]^2
To simplify the equation, let's cancel out the matching units:
[kg·m/s^2] = k [m^2] [m^2/s^2]
Now, let's simplify the right side of the equation:
[kg·m/s^2] = k [m^4/s^2]
To make the equation balanced, we need to cancel out the units on both sides:
[kg·m/s^2] = k [m^4/s^2] should be equal to just [kg·m/s^2].
To eliminate the extra m^4 on the right side, we need k to have units that cancel out the m^4. That means k should have units of m^-4.
Now, we can simplify it further:
[kg·m/s^2] = k [m^-4/s^2]
To make it simple, let's multiply k by s^2/s^2:
[kg·m/s^2] = k [m^-4/s^2] * s^2/s^2
Now, simplify the right side of the equation:
[kg·m/s^2] = k [m^-4/s^2] * (s·s)/(s·s)
[kg·m/s^2] = k [m^-4·s^2/s^2]
Finally, we can cancel out s^2 on the right side:
[kg·m/s^2] = k [m^-4·s^2/s^2] = k [m^-4]
So, the suitable unit for k is kg·m^-4, or simply kg/m^4.
However, the mark scheme answer states kg·m^-3, which seems to be a typographical error. It should be kg/m^4.
F = mass * acceleration
kg m/s^2 or Newtons = Force units
kg m/s^2 = kunits * m^2 *m^2/s^2
k units = kg /m^3