Three blocks of masses 2 kg , 3 kg , and 5 kg are connected in sequence by massless strings on a frictionless horizontal surface. A horizontal force of 40 N is applied to the 5 kg block, pulling the system forward. What is the tension in the string connecting the 2 kg and 3 kg blocks?

Step-by-step solution

The total mass of the system is m_ total = 2 kg + 3 kg + 5 kg = 10 kg . The common acceleration of the system is given by Newton's second law: a = F m_ total = 40 10 = 4 m s ⁻² To find the tension in the string connecting the 2 kg and 3 kg blocks, we isolate the trailing block ( 2 kg ). The only horizontal force acting on it is the tension T . Applying Newton's second law to the 2 kg block: T = m a = 2 4 = 8 N Answer: 8 N