>
#mathematics
,
#physics
>
Two blocks of mass 2kg (A) and 3kg (B) rest on a smooth horizontal surface and connected by a taut string of negligible mass.Force of 10N is applied to 3kg Calculate the tension in string between?
Nitha Mulungu
May 26
Two blocks of mass 2kg (A) and 3kg (B) rest on a smooth horizontal surface and connected by a taut string of negligible mass.Force of 10N is applied to 3kg Calculate the tension in string between?
1 view
1 view
2 answers
Sort By
ChatGPT
May 26
To solve this problem, we need to consider the forces acting on each block separately.
For block A (2kg):
- The only force acting on block A is the tension in the string, T.
- Since block A is not accelerating, the net force on it is zero.
- Therefore, T = 0
For block B (3kg):
- The force of 10N is applied to block B.
- The tension in the string, T, is acting in the opposite direction.
- The net force on block B is the difference between the applied force and the tension: 10N - T.
- Since block B is not accelerating, the net force on it is zero.
- Therefore, 10N - T = 0
- T = 10N
Therefore, the tension in the string between the two blocks is 10N.
For block A (2kg):
- The only force acting on block A is the tension in the string, T.
- Since block A is not accelerating, the net force on it is zero.
- Therefore, T = 0
For block B (3kg):
- The force of 10N is applied to block B.
- The tension in the string, T, is acting in the opposite direction.
- The net force on block B is the difference between the applied force and the tension: 10N - T.
- Since block B is not accelerating, the net force on it is zero.
- Therefore, 10N - T = 0
- T = 10N
Therefore, the tension in the string between the two blocks is 10N.
0
Google Gemini AI
May 26
The force applied to the 3 kg block will cause both blocks to accelerate together. The acceleration can be calculated using Newton's second law:
$$F = ma$$
where F is the net force, m is the total mass, and a is the acceleration.
In this case, the net force is 10 N and the total mass is 5 kg, so the acceleration is:
$$a = \frac{F}{m} = \frac{10 N}{5 kg} = 2 m/s^2$$
The tension in the string is equal to the force exerted by the 2 kg block on the 3 kg block. This force can be calculated using Newton's third law:
$$F_{AB} = -F_{BA}$$
where F_{AB} is the force exerted by the 2 kg block on the 3 kg block, and F_{BA} is the force exerted by the 3 kg block on the 2 kg block.
Since the 2 kg block is accelerating to the right, the force exerted by the 2 kg block on the 3 kg block must be to the left. Therefore, F_{AB} is negative.
The tension in the string is equal to the magnitude of F_{AB}, so:
$$T = |F_{AB}| = |-F_{BA}|$$
The force exerted by the 3 kg block on the 2 kg block can be calculated using Newton's second law:
$$F_{BA} = m_Ba_B$$
where m_B is the mass of the 3 kg block, and a_B is the acceleration of the 3 kg block.
The acceleration of the 3 kg block is the same as the acceleration of the 2 kg block, so:
$$F_{BA} = m_Ba_B = 3 kg \times 2 m/s^2 = 6 N$$
Therefore, the tension in the string is:
$$T = |F_{AB}| = |-F_{BA}| = |-6 N| = \boxed{6 N}$$
$$F = ma$$
where F is the net force, m is the total mass, and a is the acceleration.
In this case, the net force is 10 N and the total mass is 5 kg, so the acceleration is:
$$a = \frac{F}{m} = \frac{10 N}{5 kg} = 2 m/s^2$$
The tension in the string is equal to the force exerted by the 2 kg block on the 3 kg block. This force can be calculated using Newton's third law:
$$F_{AB} = -F_{BA}$$
where F_{AB} is the force exerted by the 2 kg block on the 3 kg block, and F_{BA} is the force exerted by the 3 kg block on the 2 kg block.
Since the 2 kg block is accelerating to the right, the force exerted by the 2 kg block on the 3 kg block must be to the left. Therefore, F_{AB} is negative.
The tension in the string is equal to the magnitude of F_{AB}, so:
$$T = |F_{AB}| = |-F_{BA}|$$
The force exerted by the 3 kg block on the 2 kg block can be calculated using Newton's second law:
$$F_{BA} = m_Ba_B$$
where m_B is the mass of the 3 kg block, and a_B is the acceleration of the 3 kg block.
The acceleration of the 3 kg block is the same as the acceleration of the 2 kg block, so:
$$F_{BA} = m_Ba_B = 3 kg \times 2 m/s^2 = 6 N$$
Therefore, the tension in the string is:
$$T = |F_{AB}| = |-F_{BA}| = |-6 N| = \boxed{6 N}$$
0
