Question

Two blocks of masses 2kg and 3kg care connected by tight string as
shown in Fig and placed on a horizontal surface given, M = 0.1 and
The acclaration of system. When the force copplied F= 45N.​

Answer 1

Answer:

velocity of center of mass

Vcm=m1×v1/m1+m2=2×5+3×(-5)/5=10-15/5=-1m/s

velocity of 2kg of mass in frame of centre of mass=5-(-1)=6m/s

Dustence of centre of mass from 2kg mass

r1=3/(2+3)×(1)=3/5m

Fc=mv^2/r1

T=2×(6)^2/3/5=2×36×5/3=120N

Therefore,tension in the string is 120 N

thank you^_^