In which case car must do more work -- Kinematics

In summary, the homework statement says that the car engine must do more work to drive from 0 kmph until it reaches 54 kmph or to reach 27 kmph from 0 kmph. The force of resistance in both cases are equal, and the relevant equations for kinetic energies are those equations: Fs = mdv / dt and v = ds / dt. However, the student does not know how to use these equations, so they will need to study kinetic energy in order to answer the question.
  • #1
students2
8
0

Homework Statement


In which case car engine must do more work: To start driving from 0 kmph until car reaches 27 kmph or to reach 54 kmph from 27 kmph? Force of resistance in both cases are equal.

I need to base answer of this problem on formulas. I apologize for any grammar mistakes (i'm not native english speaker).

Homework Equations



NA

The Attempt at a Solution


V0 = 0 kmph
27 kmph = 7,5 mps
54 kmph = 15 mps[/B]
 
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  • #2
What are the relevant equations for kinetic energies ?
 
  • #3
I think it's those equations:
1) Fs = mdv / dt
2) v = ds / dt
but I have no clue how to use them.
 
  • #4
Have you studied kinetic energy?
 
  • #5
Nope, never really
 
  • #6
students2 said:
Nope, never really

Let's make sure. The kinetic energy [itex] E [/itex] of a mass [itex] m [/itex] moving with velocity [itex] v [/itex] is [itex] E = \frac{1}{2} m v^2 [/itex].
 
  • #7
Mass isn't given but velocity is going to be higher to reach 57 kmph from 27 kmph, right?
 
  • #8
students2 said:
Mass isn't given but velocity is going to be higher to reach 57 kmph from 27 kmph, right?

Since the mass of the car stays the same, just represent it by a letter M.

What do you mean "velocity is going to be higher"?
Do you mean the "change in velocity"?
No, the change in velocity from 0 to 27 is the same as the change in velocity from 27 to 57.

But what is the change in kinetic energy from 0 to 27 versus the change in kinetic energy between 27 to 54 ?
 
  • #9
I mean, isn't velocity going to be higher when driving at 54 kmph?
So, if the velocity doesn't changes (value stays the same) and E=½mv2 is the equation, then, change in kinetic energy also is the same?
 
  • #10
students2 said:
I mean, isn't velocity going to be higher when driving at 54 kmph?
Do you mean that velocity is higher at 54 kmph than at 27 kmph? Yes, that is correct. In English is you say "higher" , you need to compare two things. It doesn't make sense to say that 54 is "higher" unless you tell about another thing..

So, if the velocity doesn't changes (value stays the same) and E=½mv2 is the equation, then, change in kinetic energy also is the same?

Same as what? It doesn't make sense to say that a single thing is "the same". You have to compare two things.
 
  • #11
Stephen Tashi said:
Do you mean that velocity is higher at 54 kmph than at 27 kmph?

Yes, exactly

But now it has confirmed that velocity will be higher while accelerating from 27 to 54 versus 0 to 27, right? So, if I put that unknown velocity value in equation as v2, change in kinetic energy (E) from 27 to 54 is also going to be higher versus 0 to 27?
 
  • #12
students2 said:
So, if I put that unknown velocity value in equation as v2, change in kinetic energy (E) from 27 to 54 is also going to be higher versus 0 to 27?

The velocity values are not "unknown".

If you compute the kinetic energy at the three known velocities, 0, 27, and 54, you can compute the change in kinetic energy beween 0 and 27 and the change in kinetic energy between 27 and 54.
 
  • #13
Why don't you do this:
Consider the three states- 1) 0 kmph 2) 27 kmph 3) 54 kmph
Find the the KE at each of these states. Then find the difference between states 2 &1 AND states 3 & 2.
What do you find?
 
  • #14
Stephen Tashi said:
If you compute the kinetic energy at the three known velocities, 0, 27, and 54, you can compute the change in kinetic energy beween 0 and 27 and the change in kinetic energy between 27 and 54.

Oh Stephen and me put up the same point I guess.
 
  • #15
Yup, i just made all calculations. (I used meters per second instead of km's per hour). I am not going to write it down, but I calculated that accelerating from 27 kmph to 54 kmph will use more KE versus 0 to 27. So, if we equalize KE to work, than car must do more work going from 27 to 54, right?
 
  • #16
students2 said:
Yup, i just made all calculations. (I used meters per second instead of km's per hour). I am not going to write it down, but I calculated that accelerating from 27 kmph to 54 kmph will use more KE versus 0 to 27. So, if we equalize KE to work, than car must do more work going from 27 to 54, right?

Correct.
 
  • #17
Yes!
 
  • #18
Thanks a lot!
 

Related to In which case car must do more work -- Kinematics

1. In which case will a car have to do more work?

The amount of work done by a car depends on the distance it travels and the force required to overcome any resistance, such as friction or air resistance. Therefore, a car will have to do more work if it travels a longer distance or encounters greater resistance.

2. How does kinematics affect the work done by a car?

Kinematics is the study of motion, and it can affect the work done by a car in several ways. For example, if the car is accelerating or decelerating, it will require more work due to the change in velocity. Additionally, if the car is moving at a constant velocity, it will only have to overcome the resistance, such as friction or air resistance, to maintain its speed.

3. How does the weight of the car impact the work it must do?

The weight of the car, or its mass, does not directly impact the work it must do. However, a heavier car will require more force to accelerate or maintain a certain speed, which in turn will require more work to be done. This is because the force needed to overcome resistance is directly proportional to the mass of the object.

4. Can the work done by a car be negative?

Yes, the work done by a car can be negative if the car is decelerating or moving in the opposite direction of the applied force. In this case, the car is actually doing work on the environment, such as slowing down or reversing direction.

5. How does the surface on which a car is driving affect the work it must do?

The surface on which a car is driving can impact the work it must do in several ways. For example, a rough surface will create more resistance, requiring the car to do more work to maintain its speed. On the other hand, a smooth surface will create less resistance, resulting in less work being done by the car. Additionally, the slope of the surface can also affect the work done by the car, as it may require more or less force to overcome the incline.

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