Question

2 bodies masses m1 and m2 are connected by a light string which passes over a frictionless massless pulley . If the pulley is moving upward with uniform acceleration g/2 , then tension in the string is -
1) 3 m1 m2 g / m1 + m2.
2)m1 + m2 g / 4m1 m2
3)2 m1 m2 g / m1 + m2
4)m1 m2 g / m1 + m2

Answer 1

⭕️When system is accelerating upward with acceleration (g/2) then the net downward acceleration. by the masses will be :-

 ⇒ g + g/2 = 3g/2

⭕️Now,

 ⇒ M(3g/2) – T = Ma➖➖➖➖(1)

 ⇒ T – m(3g/2) = ma➖➖➖➖(2)

⭕️Adding ;

 ⇒ (3g/2)(M – m) = a(M + m)

 ⇒a = (3g/2)(M – m)/(M + m)

So,

⭕️From (2) = T = m(3g/2) + ma

 ⇒∴ T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)

 ⇒∴ T = 3mgM/(M + m)

 ⇒ ∴ This is the tension in the string.

$\huge\bf{\underline{\mathfrak{thank \: you :)}}}$

Answer 2

$<b>Hello Friend<b>$ ❤️❤️

The answer of u r question is..✌️✌️


Ans:✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️


$g + \frac{g}{2} = \frac{3g}{2}$


Then,




M(3g/2)-T= Ma°°°°°°°°°°°°°°°°1

T - m(3g/2) = ma°°°°°°°°°°°°°°2



Adding equations.

(3g/2)(M-m) = a (M+m)

a=(3g/2) (M-n)/M+m


From °°°°°°°°°°°°°°°2


T = m(3g/2) (M-m) / (M+m)


T = 3mgM/(M+m)



Hope it helps uu buddy!!❤️❤️

Thank you..☺️☺️