Velocity calculation of an accelerated mass based on an increasing force

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Discussion Overview

The discussion revolves around calculating the final velocity of a mass that is being accelerated by a force that doubles over a known distance. Participants explore methods to approach the problem without knowing the time, focusing on energy considerations and the nature of the force applied.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests using energy methods to calculate the final velocity, proposing to integrate the force over the distance to find the change in kinetic energy.
  • Another participant seeks clarification on whether the doubling refers to the force or the mass, assuming it is the force that doubles based on typical physical scenarios.
  • A specific example is provided involving a variable force expressed as F(x) = 6x² + 4x, with calculations leading to a change in kinetic energy of 48 J and a resulting velocity of 4.89 m/s for a mass of 4 kg over a specified distance.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the problem statement regarding the doubling of force versus mass. There is also no agreement on a definitive method for calculating the final velocity, as different approaches are suggested.

Contextual Notes

The discussion includes assumptions about the nature of the force (linear variation) and the conditions of the problem (frictionless track, initial rest state). The lack of clarity in the problem statement regarding the doubling of force or mass introduces ambiguity in the analysis.

SpaceThoughts
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A known force is doubling (egal) over a known distance, accelerating a mass.
How do I calculate the final velocity of the mass at the end of the known distance , when the mass has doubled? I don't know the time.
The mass is accelerated from 0 meter and from 0 velocity.
 
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Since you don’t know the time, the easiest approach will be to use energy. Write down ##F(x)## then calculate ##\int F(x) \ dx## over the known distance. Assuming that is the only force then that quantity is equal to the change in KE, so you can calculate velocity.
 
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SpaceThoughts said:
Summary:: A known force is doubling (egal) over a known distance, accelerating a mass.
How do I calculate the velocity of the mass?

A known force is doubling (egal) over a known distance, accelerating a mass.
How do I calculate the final velocity of the mass at the end of the known distance , when the mass has doubled? I don't know the time.
The mass is accelerated from 0 meter and from 0 velocity.
The first step should be to clarify the problem statement and try to assign variable names for the various parameters of the problem.

At one point you say that the force is doubling over a known distance. At another point you say that the mass has doubled. Which is it? Since masses do not usually change, I will assume that it is the force that doubles.

So you have this object with mass m at rest at the left end of a frictionless track. The track has length d. The object is subject to a variable rightward force. That force varies along the length of the track. Let us denote the rightward force experienced at position x along the track as f(x). f(0) is the force experienced at the left end. f(d) is the force experienced at the right end.

We assume that the force varies linearly (because you have not told us that it is exponential). We are told that f(d) = 2f(0).

Can you find a formula for f(x)? If you like, use ##F## for the rightward force experienced at the beginning of the scenario and write your formula in terms of ##F##
 
Dale said:
Since you don’t know the time, the easiest approach will be to use energy. Write down ##F(x)## then calculate ##\int F(x) \ dx## over the known distance. Assuming that is the only force then that quantity is equal to the change in KE, so you can calculate velocity.
With variable force then is it just: ΔKE = ∫ F dx.
mass = 4 kg
between x=2 and x=3
F(x) 6x^2 + 4x
∫F(x) dx = 2x^3 +2x^2
ΔKE = 48 J
v = 4.89m/s
 
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