Calculate work done by variable force

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The discussion focuses on calculating the work done by a variable force defined as F(t) = at + b. The user presents their solution involving the integration of force and velocity over time to determine work done, and seeks confirmation of its correctness. Key points include the transformation of dr to v(t)dt and the calculation of velocity v(t) based on the force equation. Participants confirm that the mathematical approach is valid and agree on the necessity of checking the calculations. The conversation emphasizes the importance of verifying the integration process in physics problems involving variable forces.
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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