Question

Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a friction less pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.​

Answer 1

Explanation:

Resultant force= g(m1-m2)

(m1+m2) a =10×4

a=40/20

a= 2m/s^2

tension in string=g(m1+m2)

=10×20

=200 N

Answer 2

Answer

Acceleration = 2 m/s²

Tension = 96 N

Given

Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a friction less pulley

To Find

1. Acceleration of the masses
2. Tension in the string when the masses are released

How to solve

We need to write all possible equations having net forces

Solution

Let m₁ = 12 kg , m₂ = 8 kg

Find attachment for diagram .

Let's write equation for mass 1 ,

$\to \bf m_1g-T=ma\\\\\to\ \rm (12)g-T=(12)a\\\\\to\ \rm 12g-T=12a\\\\\to\ \rm 12g-12a=T\\\\\to\ \rm 12(10)-12a=T\\\\\to\ \rm 120-12a=T\\\\\to\ \bf T+12a=120...(1)$

For mass 2 ,

$\to \bf T-m_2g=ma\\\\\to\ \rm T-(8)g=(8)a\\\\\to\ \rm T=8a+8(10)\\\\\to\ \rm T=8a+80\\\\\to\ \bf T-8a=80...(2)$

Solve (1) - (2) ,

$\to \rm (T+12a)-(T-8a)=120-80\\\\\to \rm 12a+8a=40\\\\\to \rm 20a=40\\\\\to \rm{\green{a=2\ m/s^2\ \; \pink{\bigstar}}}$

On sub. a value in (1) , we get ,

$\to \rm T+12(2)=120\\\\\to \rm T+24=120\\\\\to \rm T=120-24\\\\\to \rm{ \orange{T=96\ N\ \; \pink{\bigstar}}}$