Solving a 2.1x10^3 kg Car Motion Problem

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Homework Help Overview

The problem involves a 2.1x10^3 kg car starting from rest on a sloped driveway at a 20-degree angle, with an average friction force acting against its motion. The car's speed at the bottom of the driveway is given, and the objective is to determine the length of the driveway.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the car, including friction and gravity, and how these forces affect the net work done on the car. There is an exploration of the calculations involved in determining the length of the driveway, with questions about the role of gravitational force in the net work equation.

Discussion Status

Some participants have provided guidance on considering gravitational force in addition to friction, indicating a productive direction in the discussion. There is an acknowledgment of the need to reassess the calculations based on the forces involved.

Contextual Notes

Participants note the discrepancy between their calculations and the book's answer, prompting a review of the assumptions made regarding the forces acting on the car.

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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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