Solving a Puzzling Physics Problem: Average Retarding Force

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Homework Help Overview

The discussion revolves around calculating the average retarding force acting on a 750 gram cart that slows from 4 m/s to 3 m/s over a distance of 2 meters, specifically considering air resistance as the only force at play.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for calculating the average retarding force, including kinetic energy changes and acceleration formulas. Some question the assumptions made regarding the direction of forces and signs in their calculations.

Discussion Status

Several participants have arrived at similar numerical results, though there is some variation in the sign of the force due to differing interpretations of direction. The discussion includes suggestions to verify the accuracy of the textbook answers and the possibility of errata.

Contextual Notes

Participants note the potential for errors in the textbook and discuss the importance of considering the direction of forces and velocities in their calculations. There is an acknowledgment of the need to clarify assumptions regarding acceleration and force signs.

nukengr10
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This is a fairly simple problem which i believe i know the answer to but the answer doesn't agree with the book i got it from.

there's a 750 gram cart traveling 4 m/s, 2 meters later its traveling at 3 m/s. Calculate the average retarding force assuming only air resistance.

Here what I did:

KE= (1/2)*m*v^2 so deltaKE=(.5)*(.750)*(4^2-3^2)= 2.625

The Work=f*d. Since the Change in Kinetic Energy equals the work done on the object i did:
2.625=f*2. Hence f=(2.625/2)=1.3125 N.

I think that's right but is there something i might be possibly missing?
 
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Looks good to me.
 
I think it is wanting you to calculate the average air resistance (disricbed in force) that it would take to slow a 750 gram object from 4 m/s to 3 m/s in 2 meters.
 
i tried it a slightly different way and got the same answer, only negative. i did
2aΔx=v²-u²

where u=inital velocity, v=final velocity,a=acceleration and x being displacement.
so using that i got

a= (v²-u²)/2Δx

Fnet=ma

Fnet= m ((v²-u²)/2Δx)

plug all the numbers in and you get -1.312N, negative because you are decelerating in the negative x direction.

the big thing is you must watch your sign. if you have air resistance slowing you its not going to be the same sign as your x direction and velocity. velocity implies direction as well as speed. you can make your velocity negative based on where you allocate your x and y-axis relative to your direction, but then your force would be positive. always watch the sign.
 
Last edited:
"negative because you are decelerating in the negative x direction."

sorry, that is worded poorly. it should say, because you are decelerating in the positive x direction, meaning you are accelerating in the negative direction.
 
Thanks everyone, it has been quite good to see that other people got the same answer as i did...im pretty sure the book just mistyped the answers or something
 
nukengr10 said:
Thanks everyone, it has been quite good to see that other people got the same answer as i did...im pretty sure the book just mistyped the answers or something

if it's a recent book it may have a web page with errata on it. If you come across something like this in the future it may be worth looking there first. If its not on the list you could send in the error you've found to the publishers.
 
nukengr10 said:
This is a fairly simple problem which i believe i know the answer to but the answer doesn't agree with the book i got it from.

there's a 750 gram cart traveling 4 m/s, 2 meters later its traveling at 3 m/s. Calculate the average retarding force assuming only air resistance.

Here what I did:

KE= (1/2)*m*v^2 so deltaKE=(.5)*(.750)*(4^2-3^2)= 2.625

The Work=f*d. Since the Change in Kinetic Energy equals the work done on the object i did:
2.625=f*2. Hence f=(2.625/2)=1.3125 N.

I think that's right but is there something i might be possibly missing?
I wouldn't have done it that way. At a constant acceleration (i.e. "average" acceleration), a body's "average" speed is "(initial speed+ final speed)/2". Here, the initial speed was 4 m/s and final speed was 3 m/s so its average speed during those 2 meters is (4+3)/2= 7/2 m/s. At that speed, it took 2/(7/2)= 4/7 seconds to travel 2 meters so its speed dropped from 4 m/s to 3 m/s, a difference of -1 m/s, in 4/7 seconds: the (average) acceleration was (-1)/(4/7)= -7/4= -1.75 m/s2. Since the mass is 750 g= .75 kg, the force is (.75)(-1.75)= -1.3125 N as you say.
 
Yeah i couldn't find the errata and it is a fairly new book.
 

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