Hey, this is the first time i have ever done this and i like to know how to solve it.
A car with a mass of 1350kg is tested by having it accelerated up a road with a slope of 1 in 5,increasing in speed from 2.5 to 12 m/s,covering a distance of 100m, agaisnt a frictional force of 750 N.
Determine:
The tracktive effort between driving wheels and the road surface.
The work done during period of the acceleration.
The average power developed.
My first time doing tracktive efforts and Alemberts equation, but from what i gathered on the internet:
My tutor does not seem to understand English either, we all asked for tips and guidance, he replied with 'ask your friends'.
100x1/5 =20m
1350x9.81x20= 264870.
Don't ask where the 9.81 came from, i saw quite a few answers with it in so i guessed.
Here is my attempt, possibly all wrong.
I guessed the tracktive effort is 2100.2
I got this by looking at the D'Alembert's equation:
Tracktive effort = gravitational force + inertia resistence + frictional resistance.
For Gravitational force, i put 1350 as that is mass and a force, the inertia resistence 1 over 5 as that is the resistence going up a slope and for the frictional force 750 because it says so.
Therefore 1350 + 1/5 + 750 = 2100.2
B) is work done so the equation: force x Distance = 750 x 100 = 75000
C) Average power i looked upon the internet for the equation which is: Integral power over time divided by time interval.
So just grasping at straws here: 75000 divided by 1/5 = 375000.
A car with a mass of 1350kg is tested by having it accelerated up a road with a slope of 1 in 5,increasing in speed from 2.5 to 12 m/s,covering a distance of 100m, agaisnt a frictional force of 750 N.
Determine:
The tracktive effort between driving wheels and the road surface.
The work done during period of the acceleration.
The average power developed.
My first time doing tracktive efforts and Alemberts equation, but from what i gathered on the internet:
My tutor does not seem to understand English either, we all asked for tips and guidance, he replied with 'ask your friends'.
100x1/5 =20m
1350x9.81x20= 264870.
Don't ask where the 9.81 came from, i saw quite a few answers with it in so i guessed.
Here is my attempt, possibly all wrong.
I guessed the tracktive effort is 2100.2
I got this by looking at the D'Alembert's equation:
Tracktive effort = gravitational force + inertia resistence + frictional resistance.
For Gravitational force, i put 1350 as that is mass and a force, the inertia resistence 1 over 5 as that is the resistence going up a slope and for the frictional force 750 because it says so.
Therefore 1350 + 1/5 + 750 = 2100.2
B) is work done so the equation: force x Distance = 750 x 100 = 75000
C) Average power i looked upon the internet for the equation which is: Integral power over time divided by time interval.
So just grasping at straws here: 75000 divided by 1/5 = 375000.
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Don't bother solving by the D'Alambert principle. That is just foolishness.
Contrary to D'Alambert, there is NO SUCH THING as inertial force. That is an oxymoron.
To hell with D'Alembert...learn the real Newton's second law. LET FORCES ADD UP TO NON-ZERO NET FORCE, and use that net force (dividing by the inertia) to get the acceleration.
Contrary to D'Alambert, there is NO SUCH THING as inertial force. That is an oxymoron.
To hell with D'Alembert...learn the real Newton's second law. LET FORCES ADD UP TO NON-ZERO NET FORCE, and use that net force (dividing by the inertia) to get the acceleration.
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