A 66.7-kN car is travelling at 66.5 mph when the driver decides to exit the freeway by going up a ramp. After coasting 418 m along the exit ramp the car\'s speed is 28.9 mph, and it is h = 13.1 m above the freeway. What is the magnitude of the average drag force exerted on the car?
Hint: How is the work done by a non-conservative force such as drag related to the change in the car\'s mechanical energy?
Well, if I'm going to calculate the magnitude of the average drag force, I better put on my drag racing shoes! Let's rev up those math engines! 🚗💨
Now, to calculate the magnitude of the average drag force, we can use the concept of work and mechanical energy. Work done by non-conservative forces, like drag, is equal to the change in mechanical energy.
The mechanical energy of the car is given by the sum of its kinetic energy (1/2mv^2) and its gravitational potential energy (mgh). When the car is on the freeway before going up the ramp, it has a certain mechanical energy, let's call it E1.
After coasting 418 m along the exit ramp, the car's speed decreases from 66.5 mph to 28.9 mph, and its height above the freeway increases to 13.1 m. This means that the car's mechanical energy changes to a new value, let's call it E2.
Since work done by drag is equal to the change in mechanical energy, we can calculate it as:
Work drag = E2 - E1
And the magnitude of the average drag force can be found by dividing this work by the distance traveled along the exit ramp (418 m):
Magnitude of average drag force = (E2 - E1) / 418 m
Now, all we need to do is calculate the values of E1 and E2, and plug them into the formulas! But alas, you didn't provide any information regarding the mass of the car. Without that, I'm afraid I won't be able to perform the calculation.
Well, looks like this jokester is stuck at the pit stop! Let me know if you have the mass of the car, and I'll be happy to help you calculate the average drag force.
To solve this problem, we need to use the concept of mechanical energy and work done by a non-conservative force.
The mechanical energy of an object can be defined as the sum of its potential energy and kinetic energy.
The change in mechanical energy of the car during its coasting on the ramp is equal to the work done by the drag force.
The initial mechanical energy of the car is the sum of its initial kinetic energy and its initial potential energy. The final mechanical energy is the sum of its final kinetic energy and its final potential energy.
Given:
Initial speed (v1) = 66.5 mph
Final speed (v2) = 28.9 mph
Initial potential energy (PE1) = 0 (assuming the level of the freeway is zero potential energy)
Final height (h) = 13.1 m = 42.98 ft
Acceleration due to gravity (g) = 9.8 m/s^2
First, let's convert the speeds into the same units:
66.5 mph = 29.73 m/s
28.9 mph = 12.91 m/s
The initial kinetic energy (KE1) of the car is given by the equation:
KE1 = (1/2) * mass * v1^2
We need to determine the mass of the car. The force acting on the car is the weight (mg) and its magnitude is given as 66.7 kN. We can convert this to Newton by multiplying by 1000:
Weight = 66.7 kN = 66,700 N
The weight of the car is equal to mg, where m is the mass of the car. Therefore, we can write:
66,700 N = m * 9.8 m/s^2
m = 6,806.12 kg
Now we can calculate the initial kinetic energy:
KE1 = (1/2) * 6,806.12 kg * (29.73 m/s)^2
Next, let's calculate the final potential energy (PE2) of the car using the equation:
PE2 = m * g * h
PE2 = 6,806.12 kg * 9.8 m/s^2 * 13.1 m
Now, the final kinetic energy (KE2) is given by the equation:
KE2 = (1/2) * mass * v2^2
KE2 = (1/2) * 6,806.12 kg * (12.91 m/s)^2
The change in mechanical energy (ΔE) is the difference between the final mechanical energy (ME2) and the initial mechanical energy (ME1):
ΔE = (KE2 + PE2) - (KE1 + PE1)
Finally, the magnitude of the average drag force (F) exerted on the car is equal to the negative value of the change in mechanical energy (ΔE) divided by the distance traveled (d).
F = -ΔE / d
Given:
d = 418 m
Now we can calculate the magnitude of the average drag force exerted on the car by plugging in the values into the formulas and solving the equations.
To find the magnitude of the average drag force exerted on the car, we need to calculate the work done by the drag force.
The work done by a non-conservative force, such as drag, is directly related to the change in mechanical energy. The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE).
The initial mechanical energy of the car is the sum of its kinetic energy when it is traveling at 66.5 mph and its potential energy when it is 13.1 m above the freeway.
The final mechanical energy of the car is the sum of its kinetic energy when it is traveling at 28.9 mph (after coasting along the exit ramp) and its potential energy when it is h = 13.1 m above the freeway.
First, let's calculate the initial and final kinetic energy:
Initial kinetic energy (KE1):
The initial speed of the car is given as 66.5 mph. To convert mph to m/s, we can use the conversion factor 1 mph = 0.447 m/s.
KE1 = (1/2) * mass * velocity^2
Final kinetic energy (KE2):
The final speed of the car is given as 28.9 mph. We convert this to m/s.
KE2 = (1/2) * mass * velocity^2
Next, let's calculate the initial and final potential energy:
Initial potential energy (PE1):
PE1 = mass * gravitational acceleration * height
Final potential energy (PE2):
PE2 = mass * gravitational acceleration * height
Since the height above the freeway (h) is the same in both cases, the change in potential energy (ΔPE) will be zero.
Therefore, the change in mechanical energy (ΔE) is calculated as:
ΔE = KE2 - KE1
Finally, the work done by the drag force is equal to the change in mechanical energy:
Work = -ΔE
The negative sign indicates that the work done by the drag force is negative, as it acts in the opposite direction to the motion. The magnitude of the average drag force is equal to the magnitude of the work done by the drag force.
So, to find the magnitude of the average drag force, we need to substitute the values we have into the above equations and calculate the result.
mg=66700 N, m=66700/9.8 kg, s= 418 m, v ₀=66.5 mph =29.7 m/s
v=28.9 mph=12.9 m/s, h=13.1 m
KE1-KE2= W(dr)+ PE
m•v₀²/2 - m•v²/2= F(dr) •s +m•g•h
Solve for “F(dr)”