InterviewSolution
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InterviewSolution
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Two blocks each of mass m, connected by an un-stretched spring are kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. (a)Find accelaration of the mass center. (b) Find the displacement of the centre of mass as function of time t. (c) If the extension of the spring is `X_(0)` at an instant t, find the displacements of the two blocks relative to the ground at this instant. |
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Answer» (a) Forces in vertical direction on the system are weights of the blocks and normal reaction from the ground. They balance themselves and have no net resultant. The only external force on the system is the apllied force F in the horizontal direction towards the right. `Sigma vec(F)_(i)= M vec(a)_(c ) rarr" "F=(m+m)a_(c )` `a_ (c ) = (F)/(2m)` towards right (b) The mass center moves with constant acceleration, therefore it displacement in time t is given by equation of constant accelaration motion. `x=ut+1/2at^2rarr`" "`x_c=(Ft^2)/(4m)` (c) Positions `x_(A)` and `x_(B)` of particles A and B forming a system and position `x_(C )`mass center are obtained by following eq. `Mvec(r )_(c ) = Sigmam_(i)vec(r )_(i)` Substituting values we obtain `2mx_(c )= mx_(A) + mx_(B)" "x_(c )= (x_(A) + x_(B))/(2)` Now using result obtained in part(b), we have `x_(A)+x_(B)=(Ft^(2))/(2m)` Extension in the spring at this instant is `x_(0)=x_(B)-x_(A)` From the above two equations, we have `x_(A) = 1/2((Ft^(2))/(2m)-x_(0))` and `x_(B) = 1/2((Ft^(2))/(2m) + x_(0))`. |
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