Q. 35

Find the acceleration of the 500 g block in figure (5-E18).



Answer :


Given: Refer the attachment diagram for data


m1 = 100 g = 0.1 kg , m2 = 500 g = 0.5 kg , m3 = 50 g = 0.05 kg


The free-body diagram for the system is given in attachment.


From the free-body diagram of the 500 g block,


T + 0.5a − 0.5g = 0 …..(i)


From the free-body diagram of the 50 g block,


T1 + 0.05g − 0.05a = a ….(ii)


From the free-body diagram of the 100 g block,


T1 + 0.1a − T + 0.5g = 0 ….(iii)


From equation (ii), T1 = 0.05g + 0.05a …..(iv)


From equation (i), T1 = 0.5g − 0.5a …..(v)


Equation (iii) becomes T1 + 0.1a − T + 0.05g = 0


From equations (iv) and (v), we get:


0.05g + 0.05a + 0.1a − 0.5g + 0.5a + 0.05g = 0


0.65a = 0.4 g


a=g = g= g downward


So, the acceleration of the 500 gm block is (downward).


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