InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A block of mass `m` is connected to a spring of spring constant k as shown in figure. The frame in which the block is placed is given an acceleration a towards left. Neglect friction between the block and the frame walls. The maximum velocity of the block relative to the frame is A. (a) `sqrt(m/k)`B. (b) `alphasqrt(m/k)`C. (c) `alphasqrt((m)/(2k))`D. (d) `2alphasqrt(m/k)` |
|
Answer» Correct Answer - B Solving this question relative to the frame (car) of reference. For maximum velocity (relative to frame), the block must be in equilibrium position. Let `x_0` be the equilibrium elongation in spring, then `ma=kx_0` From work-energy theorem, `(mv^2)/(2)-0=-(kx_0^2)/(2)+ma x_0` Solving the above equation, we get `v=asqrt(m/l)` This question is an application of using the work-energy theorem in non-inertial frame of reference. |
|