Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use
1 answer:
Answer:
amount of work done, W = 549.36 kJ
Given:
mass of a car engine, m = 2500 kg
initial velocity, u = 45 mph
final velocity, v = 65 mph
1 mile = 1609
Solution:
We know that 1 hour = 3600 s
Now, velocities in m/s are given as:
u = 45 mph = = 20.11 m/s
v = 65 mph = = 29.05 m/s
Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:
W =
W =
W =
W = 549.36 kJ
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Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows: