Two blocks A and B,each of the same mass,are | vedclass

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(B) For block $A$,the entire tension $T$ acts in the horizontal direction,causing horizontal acceleration $a_A = T/m$.For block $B$,the tension $T$ ac

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$t_A < t_B$

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(B) For block $A$,the entire tension $T$ acts in the horizontal direction,causing horizontal acceleration $a_A = T/m$.For block $B$,the tension $T$ acts along the string. As block $B$ swings,the angle $	heta$ between the string and the horizontal increases. The horizontal component of tension is $T \cos 	heta$. Since $\cos 	heta  0$,the effective horizontal force on $B$ is $T \cos 	heta Thus,the horizontal acceleration of $B$ is $a_B = (T \cos 	heta)/m Since both blocks cover the same horizontal distance $l$,and block $A$ has a higher average horizontal acceleration than block $B$,block $A$ will reach the pulley in less time.Therefore,$t_A < t_B$.

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