Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Answers
Answer:
Horizontal force,F=600N
mass of body A,m1=10kg
mass of body B,m2=20kg
total mass of system m=m1+m2=30kg
using Newton second load of motion, the acceleration produced by the system is calculated as
F = ma
a=F/m=600/30
=20m/s²
when force F applied on body A:
the equation of the motion can be written as
F - T= m1a
T=F - (m1)a
=600-10×30=400N (i)
when force F applied on body B:
the equation can be written as:
F - T=m2a
T=F - (m2)a
= 600 - 20×20= 200 N ( ii )
which is different from value T in case ( i).
hence our answer depends on which mass end,the force applied.
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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