Am I approach this question right? finding acceleration

In summary: I've been using the formula a = (Fnet) / m and I keep getting 7.9 m/s^2, but the actual answer is 0.61 m/s^2. I'm guessing you're right, and the book is wrong. In summary, the conversation discusses the calculation of acceleration of a sled being pulled by a tractor. The mass of the sled, applied force, and coefficient of kinetic friction are given. The discussion covers the formulas and steps used to calculate the force of friction and acceleration, with some confusion arising from the symbols used. Ultimately, it is concluded that the book may have the wrong answer as the calculated acceleration does not match the given answer.
  • #1
viet_jon
131
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.
 
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  • #2
so it should be

Fa - Ff = ma


?
 
  • #3
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.
 
  • #4
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
 
  • #5
anybody?

this question is really bothering me... it's intro physics, and I'm struggling already.
 
  • #6
Perhaps the book is wrong because I attained the same answer without reading the thread until now.
 
  • #7
lol...wow

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

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity that includes both magnitude (how much the velocity changes) and direction.

How do I find acceleration?

To find acceleration, you need to know the initial velocity, final velocity, and time interval. Then, you can use the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

What units are used to measure acceleration?

The SI unit for acceleration is meters per second squared (m/s²). However, other common units include feet per second squared (ft/s²) and kilometers per hour squared (km/h²).

What factors affect acceleration?

The main factors that affect acceleration are the mass of the object and the net force acting on it. The greater the mass, the harder it is to accelerate. The greater the force, the more the object's motion will change.

How is acceleration related to velocity and displacement?

Acceleration, velocity, and displacement are all related through the fundamental equations of kinematics. Acceleration is the derivative of velocity with respect to time, and velocity is the derivative of displacement with respect to time. This means that acceleration describes how velocity changes, and velocity describes how displacement changes.

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