Calculate work done by variable force

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Discussion Overview

The discussion revolves around calculating the work done by a variable force defined as F(t) = at + b. Participants explore the mathematical formulation and integration process involved in determining the work done over time t.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents a solution involving the integration of the force function F(t) and its relationship with velocity v(t).
  • Another participant questions whether the velocity v is zero at time t = 0, seeking clarification on the initial conditions.
  • There is a discussion about the appropriateness of substituting dr with v(t)dt in the integration process.
  • Some participants confirm that the substitution is acceptable, while others express uncertainty about the mathematical steps taken.

Areas of Agreement / Disagreement

Participants generally agree on the form of the force function and the goal of calculating work, but there is uncertainty regarding the mathematical steps and the validity of certain substitutions. No consensus is reached on the correctness of the solution presented.

Contextual Notes

Participants do not fully resolve the implications of the initial conditions or the mathematical transformations used in the integration process.

Yalanhar
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Is my solution correct?

$$ F(t) = at + b $$$$W=\int_t F(t)dr, ~~~~~~~~~ ~~dr = v(t)dt$$$$W=\int_t F(t)\cdot v(t) dt$$ $$ f = \frac{dp}{dt}$$
therefore $$v(t) = \frac{1}{m}(at^2/2+bt)$$then $$W = \int_t \frac{at+b}{m}\cdot\left(\frac{at^2}{2}+bt\right)dt$$ $$W = \frac{1}{m}\int_t \frac{a^2t^3}{2}+abt^2+\frac{abt^2}{2}+b^2tdt$$ $$W =\frac{1}{m}\left(\frac{a^2t^4}{8}+\frac{abt^3}{3}+\frac{abt^3}{6}+\frac{b^2t^2}{2}\right)$$
 
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Yalanhar said:
Summary:: I want to calculate the work done in t by a variable force that follows:
F(t) = at+b

Is my solution correct?

$$ F(t) = at + b $$$$W=\int_t F(t)dr, ~~~~~~~~~ ~~dr = v(t)dt$$$$W=\int_t F(t)\cdot v(t) dt$$ $$ f = \frac{dp}{dt}$$
therefore $$v(t) = \frac{1}{m}(at^2/2+bt)$$then $$W = \int_t \frac{at+b}{m}\cdot\left(\frac{at^2}{2}+bt\right)dt$$ $$W = \frac{1}{m}\int_t \frac{a^2t^3}{2}+abt^2+\frac{abt^2}{2}+b^2tdt$$ $$W =\frac{1}{m}\left(\frac{a^2t^4}{8}+\frac{abt^3}{3}+\frac{abt^3}{6}+\frac{b^2t^2}{2}\right)$$
is v =0 at t = 0? In addition, what is your question?
 
Chestermiller said:
is v =0 at t = 0? In addition, what is your question?
Yes
Calculate the work done by that force after time t
 
Yalanhar said:
Yes
Calculate the work done by that force after time t
So your question is whether you did the math right?
 
Chestermiller said:
So your question is whether you did the math right?
Well, yes. I don't know if can change dr to v(t)dt
 
Yalanhar said:
Well, yes. I don't know if can change dr to v(t)dt
Sure, that’s perfectly ok.
 
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Chestermiller said:
Sure, that’s perfectly ok.
Tnks
 

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