Question on Work, Energy, Power

Hi. Please help me answer this question and explain with solution. Thank you so much!

Question:
A car has a maximum output power of 20kW and a mass of 1500 kg. At what maximum velocity can it ascend an incline of 10%, assuming (a) no dissipative forces, (b) a constant resistance of 1.0 kN opposing its motion.

[Ans: 13.3 m/s, 8.0 m/s]

Reply 1

blackmarauder

5

Energy/Work Done = Force * Distance right?

Power = E/T = Force * Distance/Time which is the same as:

Power = Force * Velocity

a) The car is moving at constant velocity right? Well it needs to overcome its weight component down the plane to move up the plane. Since velocity is constant, the weight comp. down the plane = the driving force.

g = 9.81 ms^-2
wght. comp. down plane= mgsin10 = 1500gsin10

driving force = 1500gsin10

So, max. vel. from eqn [P]

V=P/F
V= [20000 W]/[1500gsin10]
V=7.8 m/s

b) Constant resistance of 1000 N has to be overcome in addition to mgsin10.

Force to be overcome= 1000 + 1500gsin10 = 3555 N

V = P / F
= 20000 / 3555
= 5.6 m/s

I would think that I am correct... but there's a high chance that it's not because I learnt this concept yesterday.

Reply 2

G. Lee

8

manfredm

Hi. Please help me answer this question and explain with solution. Thank you so much!

Question:
A car has a maximum output power of 20kW and a mass of 1500 kg. At what maximum velocity can it ascend an incline of 10%, assuming (a) no dissipative forces, (b) a constant resistance of 1.0 kN opposing its motion.

[Ans: 13.3 m/s, 8.0 m/s]

Right first of all, i know you are from singapore, so your english might not be top-notch, but what sense do you mean dissipative forces, and also, what are the relevent equations? For instance,

Power=\frac{E}{t}={W}{t}

and

F=Ma

pops into mind.

Reply 3

Willa

11

Power = P
Work done = W
Time = t

P = W/t. So W = Pt

In time t, car advances distance s=vt up the incline (v = velocity, s=distance)

For an incline of 10%, every 10m travelled horizontally you climb 1m vertically. Express this as an angle a=inverse tan(1/10).

Then when the car travels distance s along slope, it has climbed vertically distance: s*sin(a)

To a good approximation, sin(a) = 1/10, so the car climbs s/10 vertically

The change in gravitational potential energy V (that's capital V) in time t = mgh = mg*s/10 = mg*vt/10

Under no dissipative forces, all your work can be spent on increasing the gravitational potential energy, i.e. in time t, V increases by W=Pt

so Pt = mgvt/10
P = mgv/10
v = 10P/mg

P=20kW, m=1500, g=9.81
v = 13.6m/s is what I get. Close enough given my approximation I would guess

With the constant resistance of 1kN, along distance s, s*1kN of work is lost, so we subtract that from the change in V (I'll call the resistance R):

Change in V = mgvt/10 = Pt - sR = Pt - vtR

mgv/10 = P - vR

v(mg/10 + R) = P

v = P/((mg/10 + R))
= 8.0 m/s (as required)

Reply 4

G. Lee

8

You really shouldn't do the homework for the person. It's not a good way to learn. Notice how i provided him the equations instead of solving the problem for him?

Reply 5

manderlay in flames

12

G. Lee

Right first of all, i know you are from singapore, so your english might not be top-notch, but what sense do you mean dissipative forces, and also, what are the relevent equations? For instance,

Power=\frac{E}{t}={W}{t}

and

F=Ma

pops into mind.

Completely random insult to someone's language skills and then irrelevant questions yourself, you are a ****.

Reply 6

trm90

12

G. Lee

You really shouldn't do the homework for the person. It's not a good way to learn. Notice how i provided him the equations instead of solving the problem for him?

Yes, nonsensical equations such as

P=Wt

Reply 7

G. Lee

8

Typo. It was meant to be divided, obviously.

Reply 8

Willa

11

G. Lee

You really shouldn't do the homework for the person. It's not a good way to learn. Notice how i provided him the equations instead of solving the problem for him?

I work on the belief that providing a sufficient explanation of the answer provides a useful tool to learning not only the principles behind the problem, but how they should be applied. I myself have learned quite a bit of my knowledge from worked solutions. And fundamentally I support examinations being used as the major performance indicator, and nobody can help you in those - I will assume that people are mature enough to want to understand the material on which they ask questions.

Reply 9

G. Lee

8

Fair enough, valid enough point. But really, the discovery is everything from a personal viewpoin, rather than been given x amount of digits and place them all the in the right order for the person, which then reduces down to them not doing the work dependantly, not to mean however it should be done independantly. Obviously we need guidence.

Reply 10

G. Lee

8

manderlay in flames

Completely random insult to someone's language skills and then irrelevant questions yourself, you are a ****.

You think that ws insultive? I just felt that the OP may have not been able to construct the mathematical basis correctly with words. That is all.

If i was being insultive, i would have said something a great deal more harsh.

So stick your opinion where the sun don't shine.

Oh and... i was attempting to help the boy. In the future, use a little logic when deducting your ''picture'' on what is being said and how it is being said, because if i really cared not for the OP, i wouldn't have attempted to help him now would i have? Now that's an insult.

Reply 11

manfredm

OP

Hi guys. Thanks for your reply with regards to this topic. I hope this does not turn into some big showdown of who-is-better-in-physics.

I actually managed to solve the question myself. And for your information, I copied the question literally from a book that I bought from England - so any doubts cast by this question as a result of its ambiguity is of book's fault. Haha.

Thank you so much, especially to G Lee for his patience.

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