Question

Two bodies of mass 10 kg and 20 kg respectively kept on a smooth horizontal surface are tied to the end of a light string a horizontal force of 600 newton is applied to the 1. larger mass the 2.
smaller mass along the direction of string what is the tension in the string in each case draw the necessary diagram of it

Answer 1

try solving it using this formula:
F-T=ma
where F is the force
T is the tension
m is the mass
and a is the acceleration due to gravity (9.81ms^-2)

Answer 2

Answer:

Given,

Mass of body (m1) = 10 Kg

Mass of body (m2) = 20 Kg

Horizontal force F = 600 N

Let a is the acceleration of system

(A) when force F applied on A:

use free body diagram ,

For body m1 ,

F - T = m1a ______________(1)

For body m2 ,

T = m2a ___________(2)

Solve eqns (1) and (2)

a = F/(m1 + m2)

= 600/(10 + 20)

= 20 m/s²

And T = m2a = 20×20 = 400 N

(B) again, using free body diagram when force on B

For body m1 ,

T = m1a _____________(1)

For body B,

F - T = m2a ___________(2)

Solve eqns (1) and (2)

a = F/(m1 + m2)

= 600/(10+20)

= 30 m/s²

And T = 10×20 = 200 N

Explanation: