Respuesta :
Answer:
To determine the power supplied to the engine when t=5 seconds for a car with a mass of 2.8 Mg accelerating at 5.5 m/s², considering drag resistance Fw = (8.5v) N and an engine efficiency of 0.6, we need to calculate the aerodynamic drag force and power. The aerodynamic drag force is given by Fad = 0.5 * Cd * A * rho * v^2, where Cd is the drag coefficient, A is the frontal area, rho is air density, and v is velocity. The aerodynamic drag power is calculated as Pad = Fad * v.
Given the drag coefficient formula and data from sources
2
, we can compute the aerodynamic drag force and power for the car at t=5 seconds. This calculation involves determining the velocity of the car at t=5 seconds using the acceleration provided and then plugging it into the formulas mentioned above.
By applying these formulas with the known values and considering the efficiency rating of 0.6 for the engine, you can find the power supplied to the engine at t=5 seconds for this specific scenario.
Answer:
[tex]\sf 250243.125 \, \textsf{W} [/tex]
Explanation:
To determine the power supplied to the engine, we need to calculate the total force acting on the car at [tex]\sf t = 5 [/tex] seconds and then use the formula for power:
[tex]\sf \textsf{Power} = \textsf{Force} \times \textsf{Velocity} [/tex]
Given:
- [tex]\sf m_{\textsf{car}} = 2.8 \, \textsf{Mg} = 2.8 \times 10^3 \, \textsf{kg} [/tex] (mass of the car)
- [tex]\sf a = 5.5 \, \textsf{m/s}^2 [/tex] (acceleration)
- [tex]\sf F_w = 8.5v [/tex] N (drag resistance due to wind)
- Efficiency rating of the engine: [tex]\sf \textsf{Efficiency} = 0.6 [/tex]
We first need to find the velocity of the car at [tex]\sf t = 5 [/tex] seconds. Given that the car starts from rest and has constant acceleration, we can use the kinematic equation:
[tex]\sf v = u + at [/tex]
Where:
- [tex]\sf v [/tex] is the final velocity
- [tex]\sf u [/tex] is the initial velocity (which is 0 in this case)
- [tex]\sf a [/tex] is the acceleration
- [tex]\sf t [/tex] is the time
So, [tex]\sf v = 0 + (5.5 \times 5) = 27.5 \, \textsf{m/s} [/tex].
Now, we can calculate the total force acting on the car at [tex]\sf t = 5 [/tex] seconds:
[tex]\sf F_{\textsf{total}} = F_{\textsf{engine}} - F_w [/tex]
The force supplied by the engine is given by [tex]\sf F_{\textsf{engine}} = ma [/tex], where [tex]\sf m [/tex] is the mass of the car and [tex]\sf a [/tex] is the acceleration.
So, [tex]\sf F_{\textsf{engine}} = (2.8 \times 10^3 \, \textsf{kg}) \times (5.5 \, \textsf{m/s}^2) = 15400 \, \textsf{N} [/tex].
[tex]\sf F_{\textsf{total}} = 15400 - 8.5v [/tex]
At [tex]\sf t = 5 [/tex] seconds, [tex]\sf v = 27.5 \, \textsf{m/s} [/tex], so:
[tex]\sf F_{\textsf{total}} = 15400 - 8.5(27.5) [/tex]
[tex]\sf F_{\textsf{total}} = 15400 - 233.75 [/tex]
[tex]\sf F_{\textsf{total}} = 15166.25 \, \textsf{N} [/tex]
Now, we calculate the power supplied to the engine:
[tex]\sf \textsf{Power} = \textsf{Force} \times \textsf{Velocity} [/tex]
[tex]\sf \textsf{Power} = F_{\textsf{total}} \times v [/tex]
[tex]\sf \textsf{Power} = 15166.25 \times 27.5 [/tex]
[tex]\sf \textsf{Power} =417071.875 \, \textsf{W} [/tex]
Since the efficiency rating of the engine is [tex]\sf0.6[/tex], the actual power supplied to the car is [tex]\sf0.6[/tex] times the calculated power:
[tex]\sf \textsf{Actual Power} = 0.6 \times 417071.875 \\\\ = 250243.125 \, \textsf{W} [/tex]
Therefore, the power supplied to the engine when [tex]\sf t = 5 [/tex] seconds is [tex]\sf 250243.125 \, \textsf{W} [/tex].
