# Work Done/Power/Efficiency on an inclined plane

1. May 28, 2013

### Dongorgon

1. The problem statement, all variables and given/known data

A toy car of mass 20kg is pushed up the entirety of a 35m slope, inclined at an angle of arcsin(2/7) to the horizontal. The height at the top of the slope is 10m. The applied force which pushes the car up the slope is 100N, initially, the toy car is stationary.

Calculate the work done, power and efficiency of the system:

3. The attempt at a solution
For the work done, am I correct in thinking it would be the work done by the applied force, i.e.: WD=100x35=3500J?

I'd like to do this 'step by step' so to speak, any help would be much appreciated, thanks!

2. May 28, 2013

I only know how to figure out work done.
W = Fd cos θ.
θ is angle of inclination or 16.6°.

Applied force F is 100N.

d or distance should be 35m.

W = 3354.13 J

3. May 28, 2013

### 462chevelle

do you not factor in the coefficient of friction for work done?

4. May 29, 2013

We already know the applied force, so work force times displacement.
I could be wrong however...
:tongue2:

5. May 29, 2013

### milesyoung

Your calculation is correct, assuming the applied force has a direction that's parallel to the inclined plane.

Edit:
The problem statement is a bit vague with regards to 'work done by what on what' and what system you should consider when calculating the efficiency.

Last edited: May 29, 2013
6. May 29, 2013

### AlexVGheo

Here is how I would attempt at the solution [in step by step process as requested =)]:
Work Done = Force applied x Distance
= 100*35 = 3500J
(In other words I agree with you)

Power = W/t = F x V
We have F is 100N and so we must just find the V.
There is an equation which tells us that on an inclined plane V2 - U2 = 2as
V is the velocity
U is the inital velocity = 0
a is the acceleration
S =35m​
Therefore, V2=2a x 35. So if we find a we can find V and thus we can determin the power.
Now, Newton's second law gives that:
F1 + F2.... Fn = ma.
In other words the sum of all the forces is equal to mass times acceleration.
There are two forces acting: the 100N you used to push and the gravitational force of the body falling back (we will assume that there is no friction between the surface and the body)
Using trigonometry and resolving the force vector of weight we get that the gravitational force of the body down the incline plane is:
-F= -mgsinθ
=-20*2/7*9.8
= -56
We say minus because we have taken the direction of the 100N to be positive and replace the 9.8 with some other value if you have been told to use g as something else i.e. 10.
So going back to Newton's Second Law:
!00-56 = 20 x a
a =2.2ms-2
So returning to the the equation for velocity we have:
V2=2 x 2.2 x35
= 154
V = 12.41 (2 d.p.) ​
I am unsure about this step, but because the body was acceleating I think to find the power we have to take the avergae velocity which would be:
(U+V)/2 =V/2
=12.41/2
=6.2ms-1 (2.d.p)]​
So we go back to power and we have:
P = F x V
= 100 x 6.2
= 620W

Efficency = EO/EI x 100
EO = Output Energy
EI = Input Energy (Work Done)
The EO, Ibelieve, here is the energy that the body now has as gravitational potential:
EO = mgh
=20 x 9.8 x 10
=1960J
And EI is the answer to the first part of the question =3500J
So the quotient becomes:
η = EO/EI x 100
=1960/3500 x 100
= 56%

This is how I would approach the question, but it is a bit ambiguous and so I have made some assumption i.e g=9.8, friction = 0 and I took the average velocity which may not be the correct way of doing it. In any case, I hope my solution helps despite being very long =)

Last edited: May 29, 2013
7. May 29, 2013

### milesyoung

I'm sure the OP appreciates your work, but the idea is to get the OP to do the work and learn from the process. Hints are welcome if needed, solutions are not.

8. May 29, 2013

### CWatters

Is that correct?

9. May 29, 2013

### AlexVGheo

Ah, I am terrible sorry, Ill remember that in the future. Should I delete the post?

10. May 29, 2013

### AlexVGheo

No, it should be V2 - U2! I have gone back and made the adjustments, thank you =)

11. May 29, 2013

### CWatters

Dongorgon ...

Did you post the question exactly as posed or does the original question say anything about the velocity?

I ask because as Alex points out there is an issue with calculating the power. As the car gets faster the power changes. If the original question said the force pushed the car "at constant velocity" then friction would be involved and the power would be constant.

12. Jun 1, 2013

### Dongorgon

Many thanks for the replies, very much appreciated.
The question only stated that the initial velocity is 0m/s. There's a diagram also which doesn't display any frictional force, or any resistance force for that matter. I was just slightly unsure with the limited information really. Another part of the same question asks 'what are the advantages of pulling as opposed to pushing up a slope, in terms of energy loss.'
Any help on that would be great thanks!