Question: How can I calculate the power output of a car accelerating uphill?

In summary, a car of 1200kg accelerates up a 10 degree incline and accelerates from 30 km/h to 90 km/h in 8 sec. The power needed for this acceleration is 113.59 kW, assuming constant acceleration. However, the average power is 37.86 kW.
  • #1
Carrie9
10
0
Homework Statement
A car of 1200kg accelerates up a 10 degree incline and accelerates from 30 km/h to 90 km/h in 8 sec. What is the power needed
Relevant Equations
P=w/t
So my result is 102hp, but it’s not correct.
 
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  • #2
Carrie9 said:
Homework Statement:: A car of 1200kg accelerates up a 10 degree incline and accelerates from 30 km/h to 90 km/h in 8 sec. What is the power needed
Relevant Equations:: P=w/t

So my result is 102hp, but it’s not correct.
You need to show your working.
 
  • #3
a=((25-8,3)/8)=2,1 m/s^2
Fg=1200(9,8)sin10=2042,1N

Fa=ma+Fg
Fa=1200(2,1)+2042,1=4562,1N

P=Fa v
P=Fa ((Vo+Vf)/2)
=75958,965 W

=102hp
 
  • #4
Carrie9 said:
a=((25-8,3)/8)=2,1 m/s^2
Fg=1200(9,8)sin10=2042,1N

Fa=ma+Fg
Fa=1200(2,1)+2042,1=4562,1N

P=Fa v
P=Fa ((Vo+Vf)/2)
=75958,965 W

=102hp

If you have constant acceleration, do you have constant power?

Or, if you have constant power, do you have constant acceleration?
 
  • #5
PeroK said:
If you have constant acceleration, do you have constant power?

Or, if you have constant power, do you have constant acceleration?
I suppose yes
 
  • #6
Carrie9 said:
I suppose yes
##P = Fv##

As the speed increases the force decreases under constant power.
 
  • #7
Sorry but I don’t get it
 
  • #8
Carrie9 said:
Sorry but I don’t get it

What don't you get? ##F = P/v##.

If ##P## is contant and ##v## increases, then ##F## decreases.

That's why if you run, cycle or drive a car, you can't just just keep going faster and faster.
 
  • #9
I don’t get how this changes my calculations
 
  • #10
Carrie9 said:
I don’t get how this changes my calculations

It means that your calculations, which assumed constant acceleration and constant force, are wrong.

The clue is from your OP:

Carrie9 said:
Relevant Equations:: P=w/t
 
  • #11
Now I’m just confused. How do I calculate the W then? and where do I apply the forces
 
  • #12
Carrie9 said:
Now I’m just confused. How do I calculate the W then? and where do I apply the forces

Forget forces. Power is about energy.
 
  • #13
PeroK said:
Forget forces. Power is about energy.
So I need to divide the change in kinetic energy by the time? KE=(1/2)mv^2 where m is the mass (1200kg) and v is the speed (25 and 8,3 m/s)?
am I completely lost?
 
  • #14
Carrie9 said:
So I need to divide the change in kinetic energy by the time? KE=(1/2)mv^2 where m is the mass (1200kg) and v is the speed (25 and 8,3 m/s)?
am I completely lost?

Remember the car is also going uphill.
 
  • #15
So change in KE + change in PE?
PE=mg(hf-hi) where hf-hi= s sin 10
 
  • #16
Carrie9 said:
So change in KE + change in PE?
PE=mg(hf-hi) where hf-hi= s sin 10

Do you have any more information on this problem?

Either you can assume constant power; or you could assume constant acceleration and calculate the maximum power.

The second is simpler. You might try that first and see whether that gets the correct answer.
 
  • #17
PeroK said:
Do you have any more information on this problem?

Either you can assume constant power; or you could assume constant acceleration and calculate the maximum power.
No more info, got the answer 75,7W now :rolleyes:
 
  • #18
Carrie9 said:
No more info, got the answer 75,7W now :rolleyes:

Okay, that's the average power. Hmm.

The power increases linearly with speed, so the maximum required is ##113kW##.
 
  • #19
PeroK said:
Okay, that's the average power. Hmm.

The power increases linearly with speed, so the maximum required is ##113kW##.
Thank you very much for the help
 
  • #20
@Carrie9 It's great that you got to the solution!

Here's a short way to get the same result.

First, find the acceleration as you did, so:

##a = \frac{v_f - v_i}{\Delta t} = 2.083 \frac{m}{s^2}##

Now the force required to accelerate a car with acceleration ##a## on a slope of angle ##\theta = 10°## is:

##F = M \left(a + g \, \sin(\theta) \right)##

The power being delivered by the engine at the start and finish of the trip up the hill:

##P_i = F \, v_i = 37.86 \,kW##

##P_f = F \, v_f = 113.59 \,kW##
 

FAQ: Question: How can I calculate the power output of a car accelerating uphill?

What factors affect a car's power output when accelerating uphill?

The main factors that affect a car's power output when accelerating uphill include the weight of the vehicle, the incline of the road, the engine's horsepower, and the transmission's gear ratio. Other factors such as air resistance and road conditions can also play a role.

How is power output calculated for a car?

Power output is calculated by multiplying the torque (force) produced by the engine with the engine's rotational speed. This is usually measured in horsepower (hp) or kilowatts (kW). The power output can also be affected by factors such as engine efficiency and friction.

What is the difference between power and torque?

Power is the rate at which work is done, while torque is the turning force produced by the engine. In simpler terms, power determines how fast a car can accelerate, while torque determines how much weight it can tow or climb.

How does an incline affect a car's power output?

When a car is driving uphill, it needs to overcome the force of gravity pulling it down. This means that the engine needs to produce more power to maintain the same speed as it would on a flat road. The steeper the incline, the more power is needed.

Can power output be improved for a car accelerating uphill?

Yes, there are a few ways to improve a car's power output when accelerating uphill. Upgrading the engine with more horsepower, using a lower gear ratio, and reducing the weight of the vehicle can all help improve power output. However, it's important to note that these changes may also affect the car's fuel efficiency and overall performance.

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