Solving a 2.1x10^3 kg Car Motion Problem

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A 2.1x10^3 kg car starts from rest on a 20-degree slope, facing a friction force of 4.0x10^3 N, and reaches a speed of 3.8 m/s at the bottom. The initial calculations for net work done were incorrect as they did not account for gravitational force acting down the slope. The correct approach involves calculating the gravitational force component and subtracting the friction force to determine the total force. The confusion arose from overlooking the role of gravity in the motion. Ultimately, the book's answer of 5.1 m for the driveway length is confirmed as correct when all forces are considered.
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Homework Statement


a 2.1*10^3 kg car starts from rest at the top of a driveway that is sloped at an angle of 20 with the horizontal. An average friction force of 4.0*10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8m/s., What is the length of the driveway?


Homework Equations





The Attempt at a Solution



Ff=4000N
2100 Kg
X=20 degrees
Vf=3.8 m/s
Vi=0 m/s

Wnet=1/2MVf^2-1/2MVi^2
Wnet= 1/2*2000*3.8^2
Wnet= 15162 N

Wnet = FfdcosX
15162=4000cos20D
D=4.03m

but the book says 5.1m?
i can't figure out what i did wrong
thx for the help.
 
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You forgot about gravity. Friction just slows the car; it's gravity which pulls it down the slope.

To find the net work you must consider all the forces: gravity and friction. (Don't forget that they act in different directions.)
 
i see so the force applied would be Fgravity*sin20 which would give me 7045.96 and i would subtract the Ff to get the total force. right?
 
Last edited:
princesspriya said:
i see so the force applied would be Fgravity*sin20 which would give me 7045.96 and i would subtract the Ff to get the total force. right?
Right!
 
thxxx
 
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