Finding net force from acceleration, speed and distance

  • Thread starter Lis277
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  • #1
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Homework Statement


The power driving the railroad engine is switched off. The engine stops, from its speed of 16m/s^-1, without braking in a distance of 1.1km. A student hypothesis that the horizontal resistive force is constant. Based on the hypothesis, calculate the mass of the railway engine.

Homework Equations


Teacher didn't provide any relevant equations but i assume F=ma and vf^2-vi^2=2ad are important

The Attempt at a Solution


First i tried to find
acceleration

vf^2-vi^2=2ad
vf=0
vi=16
a=?
d=1.1
(0)^2-(16)^2=2(1.1)a
-256=2.2a
-116.36=a

I think the next step would be to find net force to plug all my variables into F=ma to find mass. I know net force is the total force on an object but I'm not sure how to find net force from my given variables.
 

Answers and Replies

  • #2
PeroK
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Does it seem logical to you that an object that starts at 16 m/s and stops after 1.1 km can only have one possible mass?
 
  • #3
PeroK
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PS what would be the acceleration if the train stopped in 1.1m?
 
  • #4
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Does it seem logical to you that an object that starts at 16 m/s and stops after 1.1 km can only have one possible mass?
I would assume it only has one mass
 
  • #5
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PS what would be the acceleration if the train stopped in 1.1m?
a=-116.36?
 
  • #6
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a=-116.36?
since acceleration is normally in the units of meters and i was given a distance in km, do i have to convert to m before plugging the distance in the equation to find acceleration?
 
  • #7
PeroK
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since acceleration is normally in the units of meters and i was given a distance in km, do i have to convert to m before plugging the distance in the equation to find acceleration?
What do you think?

Many distances could be 1.1 something. Inches, feet, yards, miles, light years, megaparsecs, angstrom units ...
 
  • #8
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What do you think?

Many distances could be 1.1 something. Inches, feet, yards, miles, light years, megaparsecs, angstrom units ...
acceleration would be 0.2327m/s
 
  • #9
PeroK
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acceleration would be 0.2327m/s
That looks a bit too big. How did you get that?
 
  • #10
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That looks a bit too big. How did you get that?
oops, made a mistake
vf^2-vi^2=2ad
vf=0
vi=16
a=?
d=1.1km-->1,100m
(0)^2-(16)^2=2(1100)a
-256=2200a
-0.11636m/s=a
 
  • #11
PeroK
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oops, made a mistake
vf^2-vi^2=2ad
vf=0
vi=16
a=?
d=1.1km-->1,100m
(0)^2-(16)^2=2(1100)a
-256=2200a
-0.11636m/s=a

Strictly speaking you should only really specify two significant digits and acceleration is ##m/s^2##.

On the other point, you can't know the mass without further data, as any vehicle could have this acceleration.
 
  • #12
haruspex
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you can't know the mass without further data
I suspect that the intended additional data is either the power being developed by the engine before it was switched off, or maybe the tractive force. @Lis277 , please check you were not told either of these.
 
  • #13
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I suspect that the intended additional data is either the power being developed by the engine before it was switched off, or maybe the tractive force. @Lis277 , please check you were not told either of these.
Does the fact that the train is moving at a constant velocity make any difference?
 
  • #14
jbriggs444
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Does the fact that the train is moving at a constant velocity make any difference?
The train is not moving at a constant velocity. It is slowing down.

According to Google, context for the problem can be found at: https://www.coursehero.com/file/24013128/N11-B2p1Forcespdf/

"A total horizontal resistive force of 76kN acts on the railway engine."
 

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