Question

In a frictionless pulley, 2 masses x and y hang.

Also, x > y

Find the tension on system and acceleration of system if acceleration due to gravity is g.



Right answer will be marked brainliest

Answer 1

Answer:

Calculate the tension in the rope using the following equation: T = M x A. Four example, if you are trying to find T in a basic pulley system with an attached mass of 9g accelerating upwards at 2m/s² then T = 9g x 2m/s² = 18gm/s² or 18N (newtons).

Explanation:

if the answer was helpful/useful pls mark me as the brainliest

Answer 2

According to Question, x is heavier than y ,so x will in downward direction & y in upward direction because x is greater or heavier than y .

Acceleration (a) will be be same because both are tension with same chain .

★ Let the tension is T and the weight of y is yg & weight of x is xg.

$\mathtt{ \underline{\: \: \: \: \: \: \: \: \: from \: \: \underline{ Newton \: 2 {}^{nd} \: law} \: \: \: \: \: \: \: \: \: \: }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \large \dashrightarrow\boxed{ \red{ \mathrm{f_{net} = m⋅a}}}$

Where,

• f = net force in body
• m = mass
• a = acceleration

now, using the formula for

Block - x

$\: \: \: \: \: \:\rightarrow \boxed{ \bf{xg - T = x⋅a}} - - (1)$

Block - y

$\rm{ + yg -T = y( - a)} \\ \rm{ yg -T = - ya} \\ \rm{ \cancel- ( T - yg) }= \cancel - ya\\ \: \: \: \: \: \: \rightarrow \boxed{ \bf{ T - yg = y⋅a}} - - (2)$

From equation (1) and (2)

$\: \: \: \: \: \:\: \: \: \: \: \:\rm{xg - \cancel{T} = xa}\\\: \: \: \: \: \:\: \: \: \: \: \: \rm{\cancel{T} - yg = ya}$

$\: \: \: \: \: \:\: \: \: \: \: \:$▬▬▬▬▬▬▬$\: \: \: \: \: \:\: \: \: \: \: \:$

$\rm{xg-yg=xa+ya}\\\: \: \: \: \: \: \rm {g(x-y)=a(x+y)}\\ \dashrightarrow \boxed{\mathtt{\green{a_{(acceleration)}=\frac{(x-y)g}{(x+y)}}}}$

Tension :

from equation (1)

$\mathtt{xg-T=xa}\\ \mathtt{T=xg-xa}\\ \mathtt{T=\mathtt{xg - x\frac{(x-y) g}{(x+y) }}}\\ \mathtt{T=\frac{xg(x+y)-x(x-y)g}{x+y}}\\ \mathtt{T=\frac{\cancel{x²g}+xyg\cancel{-x²g}+xyg}{x+y}}\\ \mathtt{T=\frac{xyg+xyg}{x+y}}\\ \dashrightarrow \mathtt{ \boxed{\mathtt{\green{\frac{2xyg}{x+y}}}}}$