Question

If the acceleration of wedge in the shown arrangement is $"a"m{s}^{-2}$ towards left, then at this instant, acceleration of the block (magnitude only) would be

• A. $4\alpha m{s}^{-2}$
• B. $a\sqrt{17-8\cos \alpha }m{s}^{-2}$
• C. $\left(\sqrt{17}\right)am{s}^{-2}$
• D. $\sqrt{17}\cos \frac{\alpha }{2}\times am{s}^{-2}$

Answer 1

The correct option is B $a\sqrt{17-8\cos \alpha }m{s}^{-2}$

Using the constraints, if the wedge moves by an amount 'x' towards left the total string that is released in '4x'.

and the displacement of the string along the incline of the wedge be 'y'

Or, $y=4x$ or, $\frac{d^{2}y}{d{t}^2}=4\frac{d^{2}x}{d{t}^2}$

i.e,$a_{along}=4a$

and $a_{plank}=a$

From the above figure $a_{net}=\sqrt{a_{plank}^2+a_{along}^2+2a_{along}{a}_{plank}\cos (180-\alpha )}$

or $a_{net}=a\sqrt{17-8\cos \alpha }m{s}^{-2}$