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Kailash Singh Muni
Asked: 2022-11-09T08:55:52+05:30 In:

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
image
A. (a) `sqrt(m/k)`
B. (b) `alphasqrt(m/k)`
C. (c) `alphasqrt((m)/(2k))`
D. (d) `2alphasqrt(m/k)`

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
image
A. (a) `sqrt(m/k)`
B. (b) `alphasqrt(m/k)`
C. (c) `alphasqrt((m)/(2k))`
D. (d) `2alphasqrt(m/k)`
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1 Answer

  1. 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.

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