# Am I approach this question right? finding acceleration

Am I approach this question right?......finding acceleration

1. 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?

2. Homework Equations

3. 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.

Related Introductory Physics Homework Help News on Phys.org
so it should be

Fa - Ff = ma

?

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.

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.