Calculating Power with Constant Velocity

[zaddyzad]

Homework Statement

A 1100kg car can accelerate from 0-60kph in 6.5 seconds. With constant engine power, what is the steepest hill the car could climb at 20 kph.

Homework Equations

P=W/T
Sliding force of the slope = 1100g(sinTheta)
A= 16.67/6.5 --> 2.56m/s^2
SF=MA

The Attempt at a Solution

I found the sum of the forces (1100 * 2.56) = 2816
and set the equation up as ... 2816=1100g(sinTheta)
though I know this is wrong because the answer is 23. I don't know how to incorperate that fact that the car is climbing at 20kph.

 

[gneill]

Homework Statement

A 1100kg car can accelerate from 0-60kph in 6.5 seconds. With constant engine power, what is the steepest hill the car could climb at 20 kph.

Homework Equations

P=W/T
Sliding force of the slope = 1100g(sinTheta)
A= 16.67/6.5 --> 2.56m/s^2
SF=MA

The Attempt at a Solution

I found the sum of the forces (1100 * 2.56) = 2816
and set the equation up as ... 2816=1100g(sinTheta)
though I know this is wrong because the answer is 23. I don't know how to incorperate that fact that the car is climbing at 20kph.

The first thing to establish is the power output of the engine given its performance of 0-60kph in 6.5s. (hint: what's the energy expenditure over time?)

The next bit will be a bit more tricky, establishing the power required to climb a slope at a given speed. Where will energy "go" when you climb a slope?

 

[zaddyzad]

How do I find work to find power if there are no distances given in the question ? I know I can find the sum of the forces through MA, but do I need that?

 

[gneill]

How do I find work to find power if there are no distances given in the question ? I know I can find the sum of the forces through MA, but do I need that?

Power is energy delivered per unit time. How much energy did the car gain over the given time when it went from 0 to 60 kph in 6.5 seconds?

You don't necessarily require the forces involved if you consider the energy expenditure versus time to climb the slope. Where is the energy going? If the speed along the incline is known, how much energy is being expended per unit time?

 

[zaddyzad]

It is becoming heat no? But again how would I find the energy the car has gained or has ?

 

[gneill]

It is becoming heat no? But again how would I find the energy the car has gained or has ?

You can ignore friction (so no heat). Where does kinetic energy "go" when something gains height in a gravitational field?

 

[zaddyzad]

It gains potential energy.

 

[gneill]

It gains potential energy.

Yes. So relate the speed of the car up the incline to the rate of gain of potential energy...

 

[zaddyzad]

I have no Idea how to solve this... 2816=1100g(sinTheta) is that close to how you solve it ? I know I am just missing the climbing velocity component i don't know how to incorporate.

 

[gneill]

Before you tackle the slope, have you determined the power output of the car engine? You will need to solve this portion of the problem before you can tackle the incline.

 

[zaddyzad]

the car is accelerating at 2.56m/s, and the car weighs 1100kg, if i multiply those I get the sum of the forces on the engine 2820.5N. How may that help me find the power ? How do I find the power ?

 

[gneill]

the car is accelerating at 2.56m/s, and the car weighs 1100kg, if i multiply those I get the sum of the forces on the engine 2820.5N. How may that help me find the power ? How do I find the power ?

Rather than thinking in terms of kinematic details like acceleration and force, think in terms of energy. How much energy did the car gain in going from 0 to 60 kph? In what time did it acquire this energy? So what was the power (energy/time)?

 

[zaddyzad]

so would i do 1/2mv^2... (0.5)*(1100)*(16.67)^2/6.5 ? then multiply that by 5.55 (20kph) to get the power of the car going up the slope ?

 

[gneill]

so would i do 1/2mv^2... (0.5)*(1100)*(16.67)^2/6.5 ?

Yes, that will give you the power that the car produces. The kinetic energy that the car achieves is equal to the work done, and it was done over time 6.5s.

then multiply that by 5.55 (20kph) to get the power of the car going up the slope ?

No, you got the power when you divided the change in KE by time.

As I hinted earlier, dealing with the slope is a bit trickier. There are a couple of possible approaches: (1) an "energy approach" where we use energy conservation to see at what rate the car can gain height using its power, the energy going into gravitational potential energy; (2) a force balance approach, where the force required to maintain the car's speed up the slope is considered.

I you go with the second approach, then what can you say about the net force acting on the car parallel to the slope as it climbs at a constant speed? Also, what is the relationship between power, force, and speed?

 

[zaddyzad]

so.. The power of the engine = 23'514watts, and this is constant as it climbs the hill therefor the opposing force (sliding) is = to it.

though when I set up 23514= 1100g*(Sinθ)... i still don't get the answer, I get an error.

 

[zaddyzad]

And for the relationship as force increases, so does power, and the more power there is, the greater the speed the car has.

 

[gneill]

so.. The power of the engine = 23'514watts, and this is constant as it climbs the hill therefor the opposing force (sliding) is = to it.

though when I set up 23514= 1100g*(Sinθ)... i still don't get the answer, I get an error.

That's because your 23514 number has units of power, not force. You have yet to determine what force that power delivers at 20 kph.

 

[gneill]

And for the relationship as force increases, so does power, and the more power there is, the greater the speed the car has.

There is a specific mathematical relationship that relates the power delivered by a force to an object moving at a given velocity. Hint: The unit for power is the Watt (W) which is also J/s, which is also N*m/s

 

[zaddyzad]

So I would need to divide by m and multiply by s ? How would I convert the power to N. And how do i convert 20kph to a force ?

 

[zaddyzad]

I did it

 

[zaddyzad]

Yay, thanks for the help :D

 

[zaddyzad]

Is it possible to solve this without finding power first ?

 

[gneill]

Is it possible to solve this without finding power first ?

Not that I am aware of, given the information in the problem statement.

Congratulations on getting the answer, by the way

 

[zaddyzad]

Not that I am aware of, given the information in the problem statement.

Congratulations on getting the answer, by the way

Another question of power and such. Let's assume you are only given a constant velocity can you calculate power the same way? On top of that would it be simplified to 1/2mv^2/v ----> (0.5)(M)(V) to calculate power ?

 

[zaddyzad]

I currently have a question where I have caculated the force on the tires of a car moving at a constant velocity, and it asks to find power.

 

[gneill]

Another question of power and such. Let's assume you are only given a constant velocity can you calculate power the same way? On top of that would it be simplified to 1/2mv^2/v ----> (0.5)(M)(V) to calculate power ?

The units don't work out correctly for that to be power. In fact, the units of M*V give momentum (kg*m*s^-1). The units for power are W = J/s = kg*m²/s² = N*m/s. Those are all equivalent units for power.

If something is moving at a constant speed V due to a force F being applied (perhaps it's working against friction) then the power being generated by that force is F*V. It will be the same as the power being "lost" to the friction.