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Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.?
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Three identical blocks of mass 0.6 kg are connected by light strings a...
Understanding the System
- We have three identical blocks, each with a mass of 0.6 kg.
- They are connected by light strings and move together with a uniform acceleration of 4.0 m/s².
- The total mass of the system (3 blocks) is:
Total mass = 3 × 0.6 kg = 1.8 kg.
Calculating the Total Force (F)
- According to Newton's second law, the net force acting on the system is given by:
F_net = m × a,
where:
- m = total mass = 1.8 kg,
- a = acceleration = 4.0 m/s².
- Thus,
F_net = 1.8 kg × 4.0 m/s² = 7.2 N.
- Therefore, the applied force F is:
F = 7.2 N.
Calculating Tensions in the Strings
- Let's denote the tension in the string between the first and second blocks as T1, and the tension between the second and third blocks as T2.
- For the third block (m3):
- The only force acting on it is T2.
- Using Newton's second law:
T2 = m3 × a = 0.6 kg × 4.0 m/s² = 2.4 N.
- For the second block (m2):
- The forces acting on it are T1 (pulling forward) and T2 (pulling backward):
- Thus, T1 - T2 = m2 × a.
- Substituting for m2:
T1 - 2.4 N = 0.6 kg × 4.0 m/s² = 2.4 N.
- Therefore, T1 = 2.4 N + 2.4 N = 4.8 N.
Summary
- Force F: 7.2 N
- Tension T1: 4.8 N
- Tension T2: 2.4 N
- We have three identical blocks, each with a mass of 0.6 kg.
- They are connected by light strings and move together with a uniform acceleration of 4.0 m/s².
- The total mass of the system (3 blocks) is:
Total mass = 3 × 0.6 kg = 1.8 kg.
Calculating the Total Force (F)
- According to Newton's second law, the net force acting on the system is given by:
F_net = m × a,
where:
- m = total mass = 1.8 kg,
- a = acceleration = 4.0 m/s².
- Thus,
F_net = 1.8 kg × 4.0 m/s² = 7.2 N.
- Therefore, the applied force F is:
F = 7.2 N.
Calculating Tensions in the Strings
- Let's denote the tension in the string between the first and second blocks as T1, and the tension between the second and third blocks as T2.
- For the third block (m3):
- The only force acting on it is T2.
- Using Newton's second law:
T2 = m3 × a = 0.6 kg × 4.0 m/s² = 2.4 N.
- For the second block (m2):
- The forces acting on it are T1 (pulling forward) and T2 (pulling backward):
- Thus, T1 - T2 = m2 × a.
- Substituting for m2:
T1 - 2.4 N = 0.6 kg × 4.0 m/s² = 2.4 N.
- Therefore, T1 = 2.4 N + 2.4 N = 4.8 N.
Summary
- Force F: 7.2 N
- Tension T1: 4.8 N
- Tension T2: 2.4 N
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Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.?
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Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.?.
Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three identical blocks of mass 0.6 kg are connected by light strings as shown in the figure. They move on a smootb horizontal surface with an acceleration of 4.0m/sec2 under the action of a force F. Calculate F and the tension in the two strings.?.
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