Calculate the force F(y) associated the following one-dimensional potential energies:
(a) U = - $ω$ y
(b) $U = a y^{3} - b y^{2}$
(c) $U = U_{0} s i n β y$

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Solution

The relation between force and potential energy is

$U = - \int F . d r$

$F (y) = - \frac{d U}{d y}$

(a) $U = - w y$

$F (y) = - \frac{d}{d y} (- w y)$

$F (y) = w$

(b) $U = a y^{3} - b y^{2}$

$F = - \frac{d}{d y} (a y^{3} - b y^{2})$

$F = - 3 a y^{2} + 2 b y$

$F = 2 b y - 3 a y^{2}$

(c) $U = U_{0} sin β y$

$F = - \frac{d}{d y} (U_{0} sin β y)$

$F = - U_{0} β (cos β y)$

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Calculate the force F(y) associated the following one-dimensional potential energies:(a) U = -ωy(b) U=ay3−by2(c) U=U0sinβy

The relation between force and potential energy is U=−∫F.dr F(y)=−dUdy (a) U=−wy F(y)=−ddy(−wy) F(y)=w (b) U=ay3−by2 F=−ddy(ay3−by2) F=−3ay2+2by F=2by−3ay2 (c) U=U0sinβy F=−ddy(U0sinβy) F=−U0β(cosβy)

Calculate the force F(y) associated the following one-dimensional potential energies:
(a) U = - $ω$ y
(b) $U = a y^{3} - b y^{2}$
(c) $U = U_{0} s i n β y$