Finding length of driveway (work-kinetic energy theorem)

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Homework Help Overview

The problem involves a car descending a sloped driveway, applying the work-kinetic energy theorem to find the length of the driveway. The scenario includes parameters such as the mass of the car, the angle of the slope, frictional forces, and the final speed of the car.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the work-energy principle, calculating work done against friction and gravitational forces. There are attempts to verify calculations and question the accuracy of values used, particularly the mass of the car.

Discussion Status

Some participants have provided corrections to initial calculations and are re-evaluating their results. There is an ongoing exploration of the equations used and the values substituted, with no clear consensus on the final answer yet.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for resolving discrepancies in calculations.

alyston
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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



W=ΔKE
KE=1/2MV^2
W=Fd

The Attempt at a Solution



Ff=4000N
m=2100 Kg
θ= 20°
V=3.8 m/s

W=ΔKE
W=(1/2)(2100)(3.8^2)= 14,440 J

I found the x component of gravity...
mg * sin 20
(9.81)(2000) * sin 20 = 6710 N

6710 N - 4000 N = 2710 N = net force

14440/2710 = 5.3 m

But the book says it's 5.1 m so I'm not sure what I did wrong.
Thanks!
 
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5.18 metres
 
you used 2000 kg instead of 2100
 
Oh, whoops, I didn't notice that. Thanks for pointing that out.

So I went back and redid some of my calculations:

W=(1/2)(2100)(3.8^2)= 15,162 J
(9.81)(2100) * sin 20 = 7046 N

7046 N-4000N=3046 N

15,162/3046=4.98 m

Hmmm...it's still a little off. Am I using the right equations?
 

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