As the car rolls down the hill with constant velocity, the net force on it is zero, so the resistance cancels out the gravity (they're equal in magnitude, opposite in direction). So if the F(resistance) = k*v^2, then k = F/v^2 = mg*sin@ / v^2 = 1200*(9.8)*(1/30) / (30)^2 Ns^2/m^2 = 98/225 Ns^2 / m^2
When the car travels up the hill at constant velocity, the net force on it is zero again. There are 3 forces acting on the car, gravity + resistance down the hill and the applied force (F) up the hill, which all cancel out to zero.
It's decelerating upwards, accelerating downwards. Conventionally the up-direction is taken as the "positive" direction, so that's probably why they said decelerating. Shouldn't matter too much either way, if you say it's accelerating downwards it should be fine.
Yes.
The object has two forces acting on it: gravity (mg*sin@, down the hill), and the "pulling" force (F, up the hill). Seeing as the body is accelerating up the hill with acceleration f, then the net force on the body is mf (up the hill). So:
I'll assume that "retardation" means the force opposing the truck from moving up the hill while it's climbing the hill. There are two causes of retardation, resistance (friction) and gravity. You're told resistance is 2000N (down the hill), and gravity is mg*sin@ = 200*9.8*(1/10) N = 196 N (down the hill). So the total retardation force is 2196 N (down the hill)
How so?
Not given enough information to work it out.
Yeah i got that, was wondering why they omitted the g. I shudnt think the book would be wrong at all..considering its Patel's "2nd edition" and a 4unit book etc.wogboy said:
I took up the incline as +ve. Total resistive force is 2000N. Ressitive force is the force of gravity down the incline + friction force. So i said -(mg/10) - F<sub>F</sub> = -2000.
It has to be a typo in the answers, otherwise the answer (which is supposed to be a force) contains a term m*sin@, which has the dimensions of mass, not force. mg*sin@ has the dimensions of force (through Newton's 2nd Law). The answer must somehow depend on g (it would surely be harder to push that body up the incline if you were living on Jupiter than on Earth)
I took "resistive force" to mean friction, but I could of course be wrong. What's the answer at the back of the book?
, so doing the same calculations I did before (this time with the mass of the truck being 2000 Kg),Yeah, that's true, semantics can be quite important in maths/science. The word "resistance" (in this context) is traditionally linked with friction, gravity isn't counted as a resistance (probably because gravity doesn't ALWAYS resist motion, such as when the truck is rolling downhill). The phrase "total resistance" refers to all the forces which can be classified as friction (e.g. air resistance, friction due to tyre contact with the road etc)
It would affect the net force on it, but not the net retardation force. Retardation (in this context) means how much the object is being forced against the direction it's moving. You don't need to know how much the truck was accelerating. All you need to know is that the truck was moving UP the hill at some (possibly non constant, but positive) velocity, so that the friction force is in the same direction as gravity i.e. down the hill.
Yep.
It really doesn't matter, as long as you explain yourself (e.g. write mgsin@ down the incline, rather than just mgsin@, for the final answer) then you should be fine. Having said this I'm not an expert on how the HSC exams are marked so don't sue me if you lose half a mark over it lol
