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1. Hi,
I have this question that I am stuck on.
I have done all of the questions apart from the last, which seems to be getting me stuck.

These are my following answers for the questions that I have done.
a) the Drag forces on the car must be 300N, so there is no constant acceleration due to no resultant force.
b) P = 300N x 20ms-1 = 6000W

but part c has got me stuck.
I have done the other ones.

Thanks.
Attached Images

2. Due to the hill, a new larger driving force will be needed to maintain the velocity, therefore use newtons second law with acceleration equal to 0 to find the new driving force, taking into account the cars weight and any resisitive forces. Then use P = DV to find the power.
3. (Original post by nwmyname)
Hi,
I have this question that I am stuck on.
I have done all of the questions apart from the last, which seems to be getting me stuck.

These are my following answers for the questions that I have done.
a) the Drag forces on the car must be 300N, so there is no constant acceleration due to no resultant force.
b) P = 300N x 20ms-1 = 6000W

but part c has got me stuck.
I have done the other ones.

Thanks.
Is an answer given in the book or not?
4. (Original post by derpz)
Is an answer given in the book or not?
The answer is not given in the book, no.

Update - This appears to be the answer, but I have no idea how this is derived.

Sahil_

5. (Original post by nwmyname)
The answer is not given in the book, no.

Update - This appears to be the answer, but I have no idea how this is derived.

Sahil_

I managed to work it out I think:

So we know that it requires 6000W for the first part of the question, now, if we work out the component of weight acting down the slope using trig, we get the answer of 588.2N.

Subtract this from the driving force:
300-588.2 = (-) 288.2N (more force required to travel at 20ms-1)

Now, P = fv:
P = 288.2 x 20
P = 5764

Add them up to get the resultant power:
5760 (3s.f.) + 6000 = 11700W

Not sure if this is the correct working - have a look through it and let me know what you think

EDIT: Simpler way is to just multiply the 588.2 by 20:
P = fv
P = 588.2 x 20
P = 11772
= 11700W
6. (Original post by derpz)
I managed to work it out I think:

So we know that it requires 6000W for the first part of the question, now, if we work out the component of weight acting down the slope using trig, we get the answer of 588.2N.

Subtract this from the driving force:
300-588.2 = (-) 288.2N (more force required to travel at 20ms-1)

Now, P = fv:
P = 288.2 x 20
P = 5764

Add them up to get the resultant power:
5760 (3s.f.) + 6000 = 11700W

Not sure if this is the correct working - have a look through it and let me know what you think

EDIT: Simpler way is to just multiply the 588.2 by 20:
P = fv
P = 588.2 x 20
P = 11772
= 11700W
how did you get 588.2?
7. (Original post by nwmyname)
how did you get 588.2?
Work out the angle opposide the side labelled 1 on the diagram given, which will be the same angle when you form the right angle triange to work out the componenet of weight along the slope:

Check the diagram I've made above, and it should make sense. Once we find the component acting along the slope, you know that is the downwards force, which needs to be overcome by the driving force, so it will be the same.

So we use that weight component to find out how much power is required when it travels at the same speed...20ms-1

Sorry if its a bit unclear
8. I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
Attached Images

9. (Original post by NNB_Herath)
I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
I dont see where I went wrong?

10. (Original post by NNB_Herath)
I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
both you and derpz are correct :P
There are two methods to this.
Method 1:Classic Trig approach
Spoiler:
Show

Start by finding .

Now find using , where .
. (from )
where is extra power needed.
to 3 sig figs.
Method 2: Graviatational potential gain
Spoiler:
Show
As the car is moving up the slope, it gains gravitational potential energy .
The distance the car moves up the hill in one second = 20m.
Find :

Find Vertical distance using

(from )
However as 1 second elapsed:
to 3 sig fig.
11. (Original post by The-Spartan)
both you and derpz are correct :P
There are two methods to this.
Method 1:Classic Trig approach
Spoiler:
Show

Start by finding .

Now find using , where .
. (from )
where is extra power needed.
to 3 sig figs.
Method 2: Graviatational potential gain
Spoiler:
Show
As the car is moving up the slope, it gains gravitational potential energy .
The distance the car moves up the hill in one second = 20m.
Find :

Find Vertical distance using

(from )
However as 1 second elapsed:
to 3 sig fig.
Thumbs Up !!
Good explanation ... and I did not know the 2nd method ..
THANKS !!

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