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A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?
- a)150√3 N
- b)300 N
- c)150 √2 N
- d)150 N
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 30kg block is kept on a frictionless 30° inclined plane, and it ...
- The tension in the rope can be found by the equation Fg sin30 = T
- Since the block is 30kg, this gives us a gravitational force of 300 N.
- Since the block is on a 30° slope, we multiply Fg by sin30 (½) to get a T of 150 N.
Most Upvoted Answer
A 30kg block is kept on a frictionless 30° inclined plane, and it ...
Given data:
- Mass of the block (m) = 30 kg
- Inclined angle (θ) = 30°
Free body diagram:
- The weight of the block acts vertically downwards.
- The normal force acts perpendicular to the inclined plane.
- The tension in the rope acts parallel to the inclined plane.
Analysis:
- The weight of the block can be resolved into two components: one perpendicular to the inclined plane (m*g*cosθ) and one parallel to the inclined plane (m*g*sinθ).
- The normal force cancels out the component of the weight perpendicular to the inclined plane.
- The net force acting on the block parallel to the inclined plane is m*g*sinθ.
- This net force is responsible for accelerating the block down the incline.
Calculations:
- Net force parallel to the inclined plane = m*g*sinθ
- Net force = mass * acceleration
- Therefore, m*g*sinθ = m*a
- Solving for acceleration, we get a = g*sinθ
- Substituting the values, we get a = 9.8 m/s^2 * sin(30°) = 4.9 m/s^2
Final step:
- The tension in the rope is equal to the force required to accelerate the block down the incline, which is given by Newton's second law as T = m*a.
- Substituting the values, we get T = 30 kg * 4.9 m/s^2 = 147 N ≈ 150 N
Therefore, the tension on the rope is approximately 150 N, which corresponds to option 'D'.
- Mass of the block (m) = 30 kg
- Inclined angle (θ) = 30°
Free body diagram:
- The weight of the block acts vertically downwards.
- The normal force acts perpendicular to the inclined plane.
- The tension in the rope acts parallel to the inclined plane.
Analysis:
- The weight of the block can be resolved into two components: one perpendicular to the inclined plane (m*g*cosθ) and one parallel to the inclined plane (m*g*sinθ).
- The normal force cancels out the component of the weight perpendicular to the inclined plane.
- The net force acting on the block parallel to the inclined plane is m*g*sinθ.
- This net force is responsible for accelerating the block down the incline.
Calculations:
- Net force parallel to the inclined plane = m*g*sinθ
- Net force = mass * acceleration
- Therefore, m*g*sinθ = m*a
- Solving for acceleration, we get a = g*sinθ
- Substituting the values, we get a = 9.8 m/s^2 * sin(30°) = 4.9 m/s^2
Final step:
- The tension in the rope is equal to the force required to accelerate the block down the incline, which is given by Newton's second law as T = m*a.
- Substituting the values, we get T = 30 kg * 4.9 m/s^2 = 147 N ≈ 150 N
Therefore, the tension on the rope is approximately 150 N, which corresponds to option 'D'.
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A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer? for MCAT 2025 is part of MCAT preparation. The Question and answers have been prepared according to the MCAT exam syllabus. Information about A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for MCAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer?.
A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer? for MCAT 2025 is part of MCAT preparation. The Question and answers have been prepared according to the MCAT exam syllabus. Information about A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for MCAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 30kg block is kept on a frictionless 30° inclined plane, and it is kept anchored to a wall by a rope. What is the tension on the rope?a)150√3 Nb)300 Nc)150 √2 Nd)150 NCorrect answer is option 'D'. Can you explain this answer?.
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