A car traveling upward on inclined plane

[EastWindBreaks]

Homework Statement

consider a 3300 lb car whose speed is increased by 35 mph over a distance of 200 ft while traveling up a rectilinear incline with a 15% grade. model the car as a particle, assume the tires do not slip, neglect all sources of frictional losses and drag. find the work done by the engine if the car starts from rest.

Homework Equations

w=Force*displacement* cos (theta)
k=0.5mass*vv
final velocity^2 - initial velocity^2 = 2*acceleration*displacement
potential energy = mass*g*h

The Attempt at a Solution

I solved this problem using W=PE+KE without knowing the acceleration, however,
when I try to solve this problem using " final velocity- initial velocity = 2*acceleration*displacement ", i got the wrong answer for the work.
35 mph= 51.33 ft/s ; initial velocity=0
a= 51.33/(2*200 ft) =6.5878 ft/s^2
is 35 mph not the final velocity?

it looks like my understanding of the acceleration here is incorrect, but i don't know what exactly is wrong with it, please help me~~

 

[TomHart]

Do you get the same answer for work as you did using the energy method if you just use 6.5878 for acceleration instead of (6.5878 - 4.77)?

 

[kuruman]

I would forget about acceleration and stick to the work-energy theorem. Here W_gravity + W_engine = K_final - K_initial. It will give you the work done by the engine directly.

 

[haruspex]

The equation is ΣF=ma. What is the significance of the Σ? What is it telling you to do, that you didn't do?

 

[EastWindBreaks]

The equation is ΣF=ma. What is the significance of the Σ? What is it telling you to do, that you didn't do?

thank you! it seems like i did not sum the forces up, i would get the correct answer if i sum the acceleration up: 6.5878+4.77 instead of 6.5878-4.77.
but i am still a bit confused, from what i understand, since the given increase of speed over 200 ft is a result that already contains the incline,then the acceleration i get from this speed and displacement should be the net acceleration of the car( acceleration due to the incline and acceleration due to the force of the engine), but its not, it seems like the acceleration i got here is only the acceleration due to the force of the engine, the question asks for the work done by the engine, which should contains both the work done by the incline (F1 in my diagram) and the work done by the force from the engine.

 

[EastWindBreaks]

Do you get the same answer for work as you did using the energy method if you just use 6.5878 for acceleration instead of (6.5878 - 4.77)?

thank you, i would get the answer if i sum the acceleration, but i still don't see the reason, why sum instead of subtract?this is how i see it... the net acceleration - the acceleration due to incline = the acceleration due to the engine, confused

 

[EastWindBreaks]

I would forget about acceleration and stick to the work-energy theorem. Here W_gravity + W_engine = K_final - K_initial. It will give you the work done by the engine directly.

thank you, yes, i should, but i just want to make sure i really understand the basic concept.

 

[haruspex]

6.5878+4.77

Maybe, but it might not be easy to justify.

acceleration due to the incline and acceleration due to the force of the engine

I don't know why you want to add accelerations. The vehicle's acceleration is 6.6f/s², or thereabouts. This is the result of the net force. The net force is the sum of the driving force (up slope) and component of gravity downslope, F_drive-mg sin(θ).

 

[EastWindBreaks]

Maybe, but it might not be easy to justify.

I don't know why you want to add accelerations. The vehicle's acceleration is 6.6f/s², or thereabouts. This is the result of the net force. The net force is the sum of the driving force (up slope) and component of gravity downslope, F_drive-mg sin(θ).

OMG,YES, thank you! i finally understood, i was dumb and kept thinking Fdrive= mgsine(theta)+net force... i didnt even realize Fdrive must be greater than the net force...

 

[haruspex]

kept thinking Fdrive= mgsine(theta)+net force

But that is correct. It's the same as I posted. What you used was Fdrive= net force - mgsine(theta).

 

[EastWindBreaks]

But that is correct. It's the same as I posted. What you used was Fdrive= net force - mgsine(theta).

LOL, now its getting embarrassing for me, yes, that's what i meant...the net force must be smaller than the Fdirve its 2 am here in NY...Thank you!

 

[haruspex]

LOL, now its getting embarrassing for me, yes, that's what i meant...the net force must be smaller than the Fdirve its 2 am here in NY...Thank you!

Sleep well. It's 4pm in Sydney.