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Power and The Watt Question [PLEASE HELP]

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.
Please help.

Thanks.
Screen Shot 2016-04-07 at 11.56.42.png131.4KB
Reply 1
Avatar for Sahil_
Sahil_
8
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.
Reply 2
Avatar for derpz
derpz
5
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.
Please help.

Thanks.


Is an answer given in the book or not?
Reply 3
Avatar for nwmyname
nwmyname
OP
13
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_


Screen Shot 2016-04-07 at 16.25.49.png
(edited 8 years ago)
Reply 4
Avatar for derpz
derpz
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_


Screen Shot 2016-04-07 at 16.25.49.png


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:smile:

EDIT: Simpler way is to just multiply the 588.2 by 20:
P = fv
P = 588.2 x 20
P = 11772
= 11700W
(edited 8 years ago)
Reply 5
Avatar for nwmyname
nwmyname
OP
13
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:smile:

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?
Reply 6
Avatar for derpz
derpz
5
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:

14600490364588.png

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
I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
Please check it !!
1.PNG1.8KB
Reply 8
Avatar for derpz
derpz
5
Original post by NNB_Herath
I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
Please check it !!


I dont see where I went wrong?

What answer did you get?
Original post by NNB_Herath
I guess my DIAGRAM is CORRECT....
Your diagram is WRONG as I see ....
Please check it !!

both you and @derpz are correct :P
There are two methods to this.
Method 1:Classic Trig approach

Spoiler

Method 2: Graviatational potential gain

Spoiler

(edited 8 years ago)
Original post by The-Spartan
both you and @derpz are correct :P
There are two methods to this.
Method 1:Classic Trig approach

Spoiler

Method 2: Graviatational potential gain

Spoiler


Thumbs Up !!
Good explanation ... and I did not know the 2nd method ..
THANKS !!

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