Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Find the Acceleration of the Blocks a and B in the Three Situations Shown in the Following Figure. - Physics

Sum

Find the acceleration of the blocks A and B in the three situations shown in the following figure.

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Solution

(a) 5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a             .....(i)
Again,
\[\left( \frac{1}{2} \right)T - 4g - 8a = 0\]
⇒ T − 8g − 16a = 0

From free-body diagram (2),
T = 8g + 16a                    ......(ii)

From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
\[\Rightarrow 21a = - 3g - a = - \frac{9}{7}\]
So, the acceleration of the 5 kg mass is \[\frac{9}{7} m/ s^2 \left(\text{ upward }\right)\] and that of the 4 kg mass is
\[2a = \frac{2g}{7} \left(\text{ downward }\right)\]
\[4a - \frac{T}{2} = 0\]


⇒ 8a − T = 0
⇒ T = 8a
Again, T + 5a − 5g = 0
From free body diagram-4,

8a + 5a − 5g = s0
⇒ 13a − 5g = 0

\[\Rightarrow a = \frac{5g}{13} \left(\text{ downward }\right)\]
Acceleration of mass 2 kg is \[2a = \frac{10}{13} \left( g \right)\] and 5 kg is 
\[\frac{5g}{13}\]. 

(c) T + 1a − 1g = 0
From free body diagram-5
T = 1g − 1a                        .....(i)
Again, from free body diagram-6,
\[\frac{T}{2} - 2g - 4a = 0\]
⇒ T − 4g − 8a = 0                   .....(ii)

From equation (i)
1g − 1a − 4g − 8a = 0
\[\Rightarrow a = \frac{g}{3}\left(\text{ downward }\right)\]
Acceleration of mass 1 kg is \[\frac{g}{3} \left(\text{ upward }\right)\]
Acceleration of mass 2 kg is \[\frac{2g}{3} \left(\text{ downward }\right)\] 

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 5 Newton's Laws of Motion
Q 34 | Page 82
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