Question

The force acting on a particle of mass m is given in terms of time t by 2.49. F a coS wt i + b sin wt j If the particle is initially at rest at the origin, find its (a) position and (b) velocity at any later time b (1- cos wt) j ma Ans. (a) (1- cos wt) i + (wt- sin wt) j. ma2 (b) sin wt i mu mw2

Schaum: theoretical mechanics.

please solve this. be carefull with handwriting. do not skip basic steps.

Answer 1

F = a cosuti + b sinut i atco, Ven) since F=mov at r=0 at to a cosuti + b sinut j. 22222 at = a asuti tb sinut 3 where = velocity of the particle J ao = a scos ut at + bi s sinwit at u Velocity of the c Iskowe je pacotyle Vet) - in ein wt i - boi (cost-1) p 4) = a sinwri + bi ci-cosut), mus Now since v= da , where is the position of the particle at I where al s dx = ar sinnt at + bi jo-aswidact la cosure you buis fit - Rematyti = a (1-was wt )itu ( wt - sinutss/ mw-

A = 2xy zi + 12 + xy k DXA =0 is indepoolsent of path path I j i Jadr independent of VYA =i(x²-x²) - j (2ny - 20g) Taayz r xy 1 tk (2x2 - 2x2) OOOOOOOOO AN so SA.dr is independent of path. A = v 2nyzi+nzj+ xy = ad irad je 28 comparing coefficient of i, j, kan sy x 2ny z 2 0 (my.t) = be?yz + f(4,2) a = x2 =) pemiy iz) - myz + f (2,3) = x2y & Chiy) = xyz + f(ny) take & (ny) = f (92) e f (2, 4) = 0 Ø (m1,4,4) = x?yz + c 10 = xyz + c

5А - 2мгі (?-9)j + 22 -2 ) k TXA i į k | 5 3 3 | 2 x 224 22-x ) -it-2х-2)-k(2х-о) [exA - ля i txt Т. о ..dr = path dependent