Work-Kinetic Energy Theorum Problem

  • Thread starter Thread starter ariana0923
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
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.
ariana0923
Messages
15
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top