Two blocks are in contact on a frictionless table. One has mass m and the other 2 m. A force F is applied on 2 m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be:
Acceleration of the blocks in both the case will be F/3m.
N1= m.a
N2= 2m.a
so N1/N2= 1/2
Two forces of 6N and 3N are acting on the two blocks of 2kg and 1 kg kept on frictionless floor. What is the force exerted on 2kg block by 1 kg block?
Net force is 6N-3N= 3N
Accleration of the system is 3N/3Kg = 1 m/s^2
Net Force on 1kg object is N1 - 3= 1*1 so, N1 = 4N
A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 ms-2. The tension T1 and T2 will be respectively: (use g = 9.8m/s2)
Taking both the objects as one system
T1- 20g=20a ⇒ T1= 20(g+a)=20*12= 240 N
Now taking lower object as a different system
T2-8g=8a ⇒T2=8(g+a)=8*12= 96N
Two blocks of masses 1 kg and 2 kg are connected with massless spring as shown. If the acceleration of 1 kg block is 1 m/s² towards right then the acceleration of 2 kg block is:
Spring force acting on 1 kg block is F = m.a = 1 N
Same spring force will act on 2 kg block in opposite direction. Thus net force acting on 2 kg block is 5N - 1 N = 4N
Acceleration of the block will be 4 N/2 kg = 2 m/s²