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(a) A particle is acted upon by a force vec(F) given by vec(F)=A cos omega t hat(i) + B hat(k) and its position vector vec(r)=a[(cos) ( omega t) hat(i)+ sin(omega t) hat(j)]+1/2 bt^(2)hat(k). Find the work done on the particle by the force vec(F) from time t= pi // 2omegato time t=pi //omega. (b) A particle moves under a force vec(F)= xy hat(i)+y^(2)hat(j) and traversesalong a path y=4x+1. Find the work done by the force when the particle is displaced from the point P(1, 5) to Q(2, 9). |
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Answer» Solution :(a) The POSITION vector of the particle is `vec(r)=a[COS omega t hat(i)+sin omega t hat(j)]+1/2 bt^(2)hat(k)` `:.`The velocity is given by, `vec(V)=(d vec(r))/(dt)=(-a omega sin omega t)hat(i)+(a omega cos omega t ) hat(i)+ ( bt) hat(k)` Work done `W=INT vec(F).vec(v)dt=-(a A omega)/(2) int_(pi//2 omega)^(pi//omega) sin 2 omega t dt + B b int_(pi//2omega)^(pi//omega)tdt` `=(a Aomega)/(2)xx(cos 2 omega t)/(2 omega)|_(pi//2 omega)^(pi//omega)+Bb 1/2t^(2)|_(pi//2 omega)^(pi//omega)=(a A omega)/(2)[(1)/(2 omega)-(-1)/(2 omega)]+Bb.1/2[(pi/omega)^(2)-((pi)/(2 omega))^(2)]` `=(a A omega)/(2)xx1/omega+3/8 Bb ((pi^(2))/(omega^(2)))=1/2 AA+(3 pi^(2))/(8omega^(2))Bb`. (b)Work done `W=vec(F).d vec(r)= int ( F_(x) hat(i)+F_(y)hat(j)).(dx hat(i)dy hat(j))` `= int(F_(x)dx)+int(F_(y)dy)=int xydy + int y^(2)dy=int x ( 4x+1)dx+int y^(2)dy` `=[4/3 x^(3)+(x^(2))/(2)]_(1)^(2)+[(y^(3))/(3)]_(5)^(9)=212` units. |
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