Two situations are shown in figures (a) and ( | vedclass

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(C) For figure $(a)$: The block $m_1$ is on a smooth horizontal surface and $m_2$ is hanging. The equations of motion are:$T = m_1 a_1$ ... $(i)$$m_2

Vedclass · Accepted Answer

$\frac{40}{7} \ ms^{-2}, \frac{10}{7} \ ms^{-2}$

Vedclass · Answer

(C) For figure $(a)$: The block $m_1$ is on a smooth horizontal surface and $m_2$ is hanging. The equations of motion are:$T = m_1 a_1$ ... $(i)$$m_2 g - T = m_2 a_1$ ... (ii)Adding $(i)$ and (ii),we get $m_2 g = (m_1 + m_2) a_1$,so $a_1 = \frac{m_2 g}{m_1 + m_2}$.Substituting the values: $a_1 = \frac{4 	imes 10}{3 + 4} = \frac{40}{7} \ ms^{-2}$.For figure $(b)$: This is an Atwood machine. The acceleration $a_2$ is given by:$a_2 = \left( \frac{m_2 - m_1}{m_1 + m_2} ight) g$Substituting the values: $a_2 = \left( \frac{4 - 3}{3 + 4} ight) 	imes 10 = \frac{1}{7} 	imes 10 = \frac{10}{7} \ ms^{-2}$.Thus,$a_1 = \frac{40}{7} \ ms^{-2}$ and $a_2 = \frac{10}{7} \ ms^{-2}$.

Column-$I$	Column-$II$
$(1)$ Static friction	$(a)$ Limiting friction
$(2)$ Rolling friction	$(b)$ Ball bearing
	$(c)$ Object moving on a road

Two situations are shown in figures $(a)$ and $(b)$. In each case,$m_1 = 3 \ kg$ and $m_2 = 4 \ kg$. If $a_1$ and $a_2$ are the respective accelerations of the blocks in these situations,then the values of $a_1$ and $a_2$ are respectively [ $g = 10 \ ms^{-2}$ ]

Similar Questions

Match Column-$I$ with Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Static friction $(a)$ Limiting friction

$(2)$ Rolling friction $(b)$ Ball bearing

$(c)$ Object moving on a road

$A$ block of mass $m$ moves on a horizontal rough surface with initial velocity $v$. The height of the centre of mass of the block is $h$ from the surface. Consider a point $A$ on the surface.

$A$ mass $2 \sqrt{3} \,kg$ is acted upon by two forces which are inclined to each other at $60^{\circ}$ and each of magnitude $1 \,N$. The acceleration of that mass in $SI$ system is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right]$ (in $\,m / s^{2}$)

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