Saved Bookmarks
| 1. |
Derive the work energy theorem for a variable force exerted on a body in one dimension . |
|
Answer» Solution :If a body of mass m and speed v moving in X - direction in one DIMENSIONS then its kinetic energy . `K = 1/2 mv^(2)` Intergating on both the SIDE , `(dK)/(dt) =d/(dt) (1/2 mv^(2))` ` =1/2 mxx2v . (dv)/(dt)` ` :. (dK)/(dt) =m . (dv)/(dt) xxv` ` :. (dK)/(dt) = mav [ :. (dv)/(dt) =a] ` ` :. (dK)/(dt) = F (DX)/(dt) [ :. ma = F and V = (dx)/(dt)] ` ` :. dK = F dx ` Intergating on both side from intial position `x_(i)` to final position `x_(f)` `int_(K_(i))^(K_(f)) dK = int_(x_(i))^(x_(f)) F dx` `K_(f) - K_(i) = F(x_(f)-x_(i))` ` = FDeltac ` where `Deltax = x_(f) -x_(i)` which is a work energy theorem for a variable force . Work energy theorem does notincorporate the COMPLETE dynamical information of Newton.s second law. Work energy theorem involves an integral over an interval of time , the timeinformation contained in the Newton.s second law is integrated over and is not available explicity . Newton.s second law for two or three dimensions is in vector form whereas the work energy theorem is in scalar form . |
|