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Standard XII

Physics

Work Done by Friction

Prism 1 of ma...

Question

Prism 1 of mass $m_{1}$ and with angle $α$ (shown in figure above) rests on a horizontal surface. Bar 2 of mass $m_{2}$ is placed on the prism. Assuming the friction to be negligible, the acceleration of the prism is dependent on K times $m_{1}$ by $m_{2}$ . What is the value of K?

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Solution

Let us draw the force diagram of each body, and on this basis we observe that the prism moves towards right say with an acceleration $_{1}$ and the bar 2 of mass $m_{2}$ moves down the plane with respect to 1, say with acceleration $_{21}$ , then, $_{2} =_{21} +_{1}$ (figure shown below).
Let us write Newton's second law for both bodies in projection form along positive $y_{2}$ and $x_{1}$ axes as shown in figure below.
$m_{2} g cos α - N = m_{2} w_{2 (y_{2})} = m_{2} [w_{21 (y_{2})} + w_{1 (y_{2})}] = m_{2} [0 + w_{1} sin α]$
or, $m_{2} g cos α - N = m_{2} w_{1} sin α . . . . (1)$
and $N sin α = m_{1} w_{1} . . . . . . . (2)$
Solving (1) and (2), we get
$w_{1} = \frac{m_{2} g sin α cos α}{m_{1} + m_{2} {sin}^{2} α} = \frac{g sin α cos α}{(\frac{m_{1}}{m_{2}}) + {sin}^{2} α}$
$∴ k = 1$

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Prism 1 of mass m1 and with angle α (shown in figure above) rests on a horizontal surface. Bar 2 of mass m2 is placed on the prism. Assuming the friction to be negligible, the acceleration of the prism is dependent on K times m1 by m2 . What is the value of K?

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