Work-Kinetic Energy Theorum Problem

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The problem involves a car with a mass of 2.1 x 10^3 kg starting from rest on a 20-degree slope, facing a friction force of 4.0 x 10^3 N, and reaching a speed of 3.8 m/s at the bottom. The net work done on the car is equal to the change in kinetic energy, which was calculated but led to a discrepancy with the book's answer of 5.1 m for the driveway length. The discussion highlights the importance of considering gravitational potential energy, which contributes positively to the car's kinetic energy as it descends the slope. It emphasizes that the gravitational force does positive work while friction does negative work, affecting the overall energy balance. Understanding these forces is crucial for accurately solving the problem.
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Homework Statement



A 2.1 x 10^3 kg car starts from rest at the top of a driveway that's sloped at an angle of 20.0 degrees with the horizontal. An avg. friction force of 4.0 x 10^3 impedes the cars motion so that the cars speed at the bottom of the driveway is 3.8m/s. What is the length of the driveway?


Homework Equations


Wnet = change in KE
KE=1/2mv^2
(so you can substitute it)
Wnet = Fdcos(angle)



The Attempt at a Solution



I'm racking my brain here! I don't know what I'm doing wrong...

Wnet = (Fnet)(d)(cos angle)
Wnet = change in KE
Wnet = (KE f) - (KE i)
Wnet = KEf - 0
Wnet = 1/2mvf^2

*****note mvf is mass(final velocity)^2*****

1/2mvf^2 = Fdcos(angle)
1/2(2.1 x 10^3)(14.44) = 4.0 x 10^3(cos 20) (d)
15,162 = 3758.77d
d=4.03 m

**The book's answer is 5.1 m

Now, I'm not sure how to handle the fact that friction is acting in the opposite direction that the work is being done, so i don't know how to account for that negative work, since the answer is not negative.

But other than that, i don't know how to get 5.1

Any help is appreciated!
 
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The car starts from the top of a slope... that means there's some gravitational potential energy you haven't included.
 
physics girl phd said:
The car starts from the top of a slope... that means there's some gravitational potential energy you haven't included.

That can't be though...I wrote everything the problem said. Plus, this section does not cover grav. potential energy, so it cannot be in the problem (the next section talks about it).
 
Regardless of where it's discussed in your text, you have to include the effect of the gravitational force, which is doing work on the object to change some of the energy of the object from potential energy to kinetic energy.

If you decide to treat it as a force... draw a free body diagram of the object on a slope. there should be three forces, the gravitational force, the friction force, and the support force of the slope (the normal force).

1) Gravitational force does work on the object to put energy in (in a certain manner dependent on the slope)... let's call that POSITIVE work.
2) The friction force takes energy away from the object (NEGATIVE work).
3) The final energy of the object (it's kinetic energy) is the positive work done on the object minus the negative work.

Thats a way of doing it without the definition of gravitational potential energy.
 
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