Calculate work done by variable force

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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
 
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
 
Chestermiller said:
Sure, that’s perfectly ok.
Tnks