Question

It's quiet in the physics section: Acceleration problem: A car of weight 7000n travels at a steady speed 8m/s up a steady incline at 15 degrees above the horizontal. The car is acting against a frictional force of 500n.

Answer 1

A car of weight 7000n travels at a steady speed 8m/s up a steady incline at 15 degrees above the horizontal. The car is acting against a frictional force of 500n.

This is a two part question, I've already solved the car's engine power = 18.5kw

The second part reads: If the car traveled down the incline with the same engine thrust and friction, what would its acceleration be?

I'm not looking for the answer, but a nudge in the right direction would be appreciated.

Thanks in advance!

Answer 2

I am trying to figure out how or where the engine power fits into an equation. Can you show how you calculated it?

Obviously, the acceleration up the hill is zero, so the engine is compensating for gravity and friction.

Since the friction is the same going up and down the hill, you only have to calculate the increasing speed or acceleration do to gravity at the given angle.

Answer 3

I calculated it by taking the energy required for the car to move 8m/s up the incline and adding it to the work done against friction so:

8 * sin 15 = height/s = 2.07 m

2.07 * 10 * 700kg = 14.5kw

500 * 8m/s = 4kw

14.5 + 4kw = 18.5kw

I assumed that since the car had the same engine thrust, it would somehow increase its acceleration down the hill?

Answer 4

@Michele_Laino

I'm at a bit of a loss here, sorry if the ping disturbs you! ;n;

Answer 5

No problem. This is a very intriguing question. I need to study your work ....

Answer 6

The real problem is that I have the engine in watts rather than joules, it would accelerate down the slope due to gravity and it's engine power. The only hindering force is the friction.

Answer 7

when the car travels down the incline, the force acting on it has the subsequent magnitude:
\[\Large F = mg\sin 15 - R = 1811.73 - 500 = 1311.73\;newtons\]

Answer 8

Your units are confusing me. Or maybe the 10 through me but that is the 9.8 m/s^2 in F = mg

8 * sin 15 = height/s = 2.07 m 8 m/s sin 15 = 2.07 m/s y velocity
2.07 * 10 * 700kg = 14.5kw (2.07 m/s)(9.8 m/s^2)(700 kg) = 14.2 KW
Your right i didn't recall W = kg m^2/s^3
500 * 8m/s = 4kw
14.5 + 4kw = 18.5kw

Answer 9

Yes, that's correct retirEEd! (Also, my book told me to take gravity as 10m/s )

If the magnitude of force acting on the vehicle is 1311.73, (Thanks, Michele!) I still need to combine it with the engine thrust somehow and then plug the values into F = ma, and that should do it!

Answer 10

the car also can travel the incline down with the motor off, in such case the acceleration will be:
\[\Large a = \frac{F}{m} = \frac{{1311.73}}{{7000/9.81}} = ...?\]

Answer 11

I am not getting this. so i is time to clear my head and go for a walk with my dog. Sorry and good luck. I might think of something while walking. I will be back in a couple of hours.

Answer 12

I don't think they want me to assume that the motor is off, the answer they gave me is 5.17 m/s

Answer 13

You and me both retirEED! Have a nice walk with your dog! c:

Answer 14

Kinetic energy formula probably won't work, since since it requires velocity, not acceleration.

Answer 15

I'm thinking...

Answer 16

What if we assume that the engine thrust they are referring to is not the wattage but the initial speed?

\[(u + v)t/2 = s\]

Assuming 1 second passes:

\[8 + (8+a)/2 = 8 + a/2\]

\[v^2 = u^2 +2as\]

\[(8+a)^2 = 80 + a\]

Answer 17

\[64 + 2a + a^2 = 80+a = a^2 +a - 24\]

Answer 18

Not really, but it's the closest answer I have so far!

Answer 19

here is my reasoning:
if the power developed by the car is \(18kW\) at \(8\) meters/second, then the average force developed by the car, is:
\[\Large F = \frac{{power}}{{speed}} = \frac{{18500}}{8} = 2,312.5\;newtons\]
now, when the car travels down the incline, then on it are acting three forces:
1) the force developed by the motor \(F=2312.5\) newtons
2) gravity \(mg \sin 15=1811.73\) newtons
3) friction \(500\) newtons in opposite direction
so total force is:
\[\Large T = 2312.5 + 1811.73 - 500 = 3,624.23\;newtons\]
and the acceleration, will be:
\[\Large a = \frac{T}{m} = \frac{{3624.23}}{{7000/10}} \simeq 5.177\]
where I used this value for gravity: \(g=10\) \(m/sec^2\)

Answer 20

That's it! I wasn't aware of the power/speed = F, (Probably forgot it, haha.) If you know the power + speed of an object, I suppose this would be a reliable way of finding out the force acting on it.

Thank you! You've been a great help. c:

Answer 21

:)

Answer 22

Food for thought.....

Hey I see the problem is closed, so you must an answer. But while I was walking, the only difference is up the hill and down the hill, right?

So with that reasoning, I'm fighting 9.8 sin(15) m/s^ going up, so going down the acceleration will be double the effect of gravity.

gravity + the engine acceleration from fighting gravity = 2 (9.8) sin(15)

It just seems logical to me.

Have a good day.