Question

three blocks of masses 3kg,4kg,5kg are connected to each other with light strings and are then placed on a smooth frictionless surface . let thr system be pulled with a force F from the side of lighter mass so that it moves with an acceleration of 1m/s². then the value of F is​

Answer 1

Answer:

12 N

F1 = m1a = 3

F2 =m2a = 4

F3 = m3a = 5

total Force F = 12 N

Answer 2

The value of the external force F is equal to 12 Newton.

Given:

Mass of the first block (m₁) = 3 kg

Mass of the second block (m₂) = 4 kg

Mass of the third block (m₃) = 5 kg

Acceleration of the system (a) = 1 ms⁻²

To Find:

The value of the force (F).

Solution:

→ The acceleration of a particle is equal to the change in velocity of the particle per unit time. It can be said that acceleration is equal to the rate of change of velocity.

→ According to Newton's second law of motion, the acceleration of a body is equal to the Net external force applied per unit mass. Thus we can calculate the acceleration (a) of a system by dividing the Net external force (F) by the total mass of the system (M).

                       $Acceleration (a)=\frac{Net ExternalForce(F)}{Mass(M)}$

→ For the given question, the total mass of the system would be equal to the sum of the individual masses of the three blocks:

                  ∴  Total Mass (M) = m₁ + m₂ + m₃

                                         ∴ M = 3 + 4 + 5

                                         ∴ M = 12 kg

→ Now the net external force (F) will be equal to the product of the acceleration (a) of the system and the total mass (M) of the system.

                  ∴ Force (F) = Total Mass (M) × Acceleration (a)

                               ∴ F = M × a

                               ∴ F = (12) × (1) = 12 N

Hence the value of the external force F comes out to be equal to 12 N.

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