How is the minimum work needed to push a car up an inclined plane calculated?

[Abu]

Homework Statement

What is the minimum work needed to push a 1000-kg car 300 m up a 17.5 degree incline?
A. Ignore friction
B. Assume the coefficient of friction is 0.25

Homework Equations

W = F*d
Net force = ma

The Attempt at a Solution

At first I did:
Work = F*300*cos17.6
but I later realized that the force and distance are both in the same direction, thus the angle would be 0
So that means that the new formula is Work = ma*300... but what is a? Acceleration is equal to force applied minus force of friction divided by mass, but the force applied is not known, and for part A friction is not included.

So I've already looked at other solutions and I found the answer, but I don't understand why it is done this way. Basically the minimum work is equal to mg*300sin17.5.

My question is how come this equation is used instead? Is the work that it takes to lift the car straight upwards the height of the incline equal to the work that it takes to push the car all the way up the inclined, and if so, how come?

Thanks.

 

[andrewkirk]

So that means that the new formula is Work = ma*300... but what is a?

It's better to forget about acceleration and just use the forces. The car experiences a downward force equal to its weight, which is mg. Then we need to draw a vector diagram and split that vertical downwards force into two components, one parallel to the slope and one perpendicular (normal) to it. The normal component is canceled out by the slope pushing back against the car, leaving the parallel force, which is the one you need to multiply by the distance traveled along the slope.

Is the work that it takes to lift the car straight upwards the height of the incline equal to the work that it takes to push the car all the way up the inclined, and if so, how come?

In part A the answer is Yes, because we ignore friction and air resistance and those are the only differences between the work in the two approaches. In part (b) the answer will be No. But I don't know what they expect for part B since rolling motion - which is what the car is doing - is much more complex than sliding motion, so more is needed than just a coefficient of friction. They should have made it a sled rather than a car.

 

[haruspex]

Work = ma*300

That equation is wrong.
The work done by a given force F_app is F_app*d cos(θ), while the acceleration a satisfies F_net=ΣF=ma. The applied force F_app is only one of the contributors to ΣF.
In this problem, you do not need the vehicle to have any residual velocity at the top, so the acceleration can be made arbitrarily small. It becomes, almost, a statics problem, ΣF=0. So what does F_app equal? (Different answers for parts A and B.)

 

[Abu]

That equation is wrong.
The work done by a given force F_app is F_app*d cos(θ), while the acceleration a satisfies F_net=ΣF=ma. The applied force F_app is only one of the contributors to ΣF.
In this problem, you do not need the vehicle to have any residual velocity at the top, so the acceleration can be made arbitrarily small. It becomes, almost, a statics problem, ΣF=0. So what does F_app equal? (Different answers for parts A and B.)

Oh okay, so let me see if I got this right . Since it is asking for the minimum work, thus the minimum force needed, the net force is zero because any extra force will only contribute to making the car move faster.

So for net force to be zero, the force applied must be equal to the resistant force, which is the component of gravity acting along the incline, which is mgsintheta. so that means force applied equals mgsintheta, and when multiplied by the distance, the formula is Work = mgsin17.5*300.

Am I correct? Thank you.

 

[haruspex]

Oh okay, so let me see if I got this right . Since it is asking for the minimum work, thus the minimum force needed, the net force is zero because any extra force will only contribute to making the car move faster.

So for net force to be zero, the force applied must be equal to the resistant force, which is the component of gravity acting along the incline, which is mgsintheta. so that means force applied equals mgsintheta, and when multiplied by the distance, the formula is Work = mgsin17.5*300.

Am I correct? Thank you.

Yes.