Am I doing my homework correctly?
This is the problem:
1) A car with a mass of 1500kg accelerates uniformly at 6m/s^2 for a distance of 100m along a horizontal road. Friction (between the tires and road) applies a constant force of 600N for the duration of the trip. Ignore wind resistance.
It asks to draw all the forces acting on the car
so i drew a + shape with Force of the car pointing right,Force of friction pointing left, and Normal force pointing up and Gravity pointing down.
For the second part, it asks for me to calculate the work done by every force.
For the force of the car i did
I used the equation w=F*d*cos@
w=(1500kg)(6m/s^2)(100m)(cos0)
and got w=0J
I got the same thing for friction
w=(600N)(100m)(cos0)
w=0J
Now I am confused with the remaining forces: normal force and gravity.
I'm confused on how I can get the angles to plug into the equation.
This is all I could plug in....I'm not sure about the angles and distance w=(1500kg)(9.8m/s^2)(0)(cos?)
My question is how do I figure out the angles and distance to find the work !
This is the problem:
1) A car with a mass of 1500kg accelerates uniformly at 6m/s^2 for a distance of 100m along a horizontal road. Friction (between the tires and road) applies a constant force of 600N for the duration of the trip. Ignore wind resistance.
It asks to draw all the forces acting on the car
so i drew a + shape with Force of the car pointing right,Force of friction pointing left, and Normal force pointing up and Gravity pointing down.
For the second part, it asks for me to calculate the work done by every force.
For the force of the car i did
I used the equation w=F*d*cos@
w=(1500kg)(6m/s^2)(100m)(cos0)
and got w=0J
I got the same thing for friction
w=(600N)(100m)(cos0)
w=0J
Now I am confused with the remaining forces: normal force and gravity.
I'm confused on how I can get the angles to plug into the equation.
This is all I could plug in....I'm not sure about the angles and distance w=(1500kg)(9.8m/s^2)(0)(cos?)
My question is how do I figure out the angles and distance to find the work !
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Force = (ma), = 1500 x 6, = 9,000N.
Distance = 100m.
Work done by car = (fd), = 9,000 x 100, = 900,000 Joules.
Force = 600N.
Distance = 100m.
Work done by friction = (fd) = 600 x 100, = 60,000N.
Net work done by car = (900,000 - 60,000) = 840,000 Joules.
Normal force = (mg) = 1500 x 9.8, = 14,700N. No work is done.
Gravitational force = (mg) = 14,700N. No work is done.
Net vertical forces = 0.
Distance = 100m.
Work done by car = (fd), = 9,000 x 100, = 900,000 Joules.
Force = 600N.
Distance = 100m.
Work done by friction = (fd) = 600 x 100, = 60,000N.
Net work done by car = (900,000 - 60,000) = 840,000 Joules.
Normal force = (mg) = 1500 x 9.8, = 14,700N. No work is done.
Gravitational force = (mg) = 14,700N. No work is done.
Net vertical forces = 0.
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Work done by the vehicle:
W = F * D
W = (1500 kg)(6 m/s^2)(100 m)(cos(0)
W = 900,000 J BTW(cos(0) = 1) (sin(0) = 0)
Work done by friction:
W = -(600 N)(100 m)(cos(0))
W = -60,000 J
Work done by the normal force:
Normal force is not applied over a distance so the work done by the force will always be 0 J.
Work done by gravity:
Will be 0 J in this case due to the fact gravity is not being applied in moving this particular vehicle since it is moving horizontally. You would have to take the work done by gravity into consideration if for example the car was falling from a fixed height.
W = F * D
W = (1500 kg)(6 m/s^2)(100 m)(cos(0)
W = 900,000 J BTW(cos(0) = 1) (sin(0) = 0)
Work done by friction:
W = -(600 N)(100 m)(cos(0))
W = -60,000 J
Work done by the normal force:
Normal force is not applied over a distance so the work done by the force will always be 0 J.
Work done by gravity:
Will be 0 J in this case due to the fact gravity is not being applied in moving this particular vehicle since it is moving horizontally. You would have to take the work done by gravity into consideration if for example the car was falling from a fixed height.