Am I approach this question right? finding acceleration

Click For Summary
The discussion revolves around calculating the acceleration of a sled being pulled by a tractor, factoring in the force applied and the frictional force. The applied force is 1.3 kN, and the sled's mass is 11,000 kg, with a kinetic friction coefficient of 0.8. Initial calculations incorrectly included friction as an additive force, leading to an incorrect acceleration of 7.9 m/s². The correct approach involves subtracting the frictional force from the applied force, resulting in an acceleration of approximately -7.73 m/s², indicating deceleration. However, confusion arises as the textbook states the answer should be 0.61 m/s², leaving participants questioning the accuracy of their calculations or the textbook itself.
viet_jon
Messages
131
Reaction score
0
Am I approach this question right?...finding acceleration

Homework Statement



A tractor applies a force of 1.3kN to the sled, which has a mass of 1.1x10^4 kg. At that point, the co-efficient of kinetic friction between the sled and the ground has increased to .8. What is the acceleration of the sled?


Homework Equations




The Attempt at a Solution



Given


M = 11,000kg
Applied Force = 1300 N , let if be Fa
Kinetic Friction co-efficient = 0.8 , let it be uK



Legend


let Fa = applied force
let Ff = force of friction
m = mass
a = acceleration


Finding force of friction


Ff = uk*Fn
= (0.8)* (11000 kg x 9.81m/s^2)
= 86 328 N



Calculating Acceleration


F=ma
Fa + Ff = ma
a = ( Fa + Ff ) / m
= ( 1300 N + 86 328 N) / 11 000 kg
= 7.9 m/s^2




I tried to do this question literally 10 times now. I keep getting this answer, but know it's wrong. Maybe my approach isn't correct, or I am missing a negative sign somewhere. Please help.
 
Physics news on Phys.org
so it should be

Fa - Ff = ma


?
 
viet_jon said:
so it should be

Fa - Ff = ma


?
yes
Applied force(Fa) - frictional force(Ff) = mass times acceleration

but don't confuse the 'a' in 'Fa' with acceleration
- its just being used to denote 'applied', the symbols used are a little confusing and they should have made more effort to avoid that.
 
YellowTaxi said:
yes
Applied force(Fa) - frictional force(Ff) = mass times acceleration

but don't confuse the 'a' in 'Fa' with acceleration
- its just being used to denote 'applied', the symbols used are a little confusing and they should have made more effort to avoid that.



Fa - Ff = ma

1300 N - 86328 N = m * a

therefore acceleration = (1300N - 86328N) / 11 000 kg

= -7.729 m/s^2



but in the book, the answer is 0.61 m/s^2
 
anybody?

this question is really bothering me... it's intro physics, and I'm struggling already.
 
Perhaps the book is wrong because I attained the same answer without reading the thread until now.
 
lol...wow

I've been beating my self all day over this question.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K