NEET Exam > NEET Questions > A block of mass 1 kg is at rest on a horizont... Start Learning for Free
A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60∘ from the horizontal that will just start the block moving is:
- a)5 N
- b)5.36 N
- c)74.6 N
- d)10 N
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A block of mass 1 kg is at rest on a horizontal table. The coefficient...
Understanding the Problem
To find the force required to just start moving a block on a table, we need to consider the forces acting on the block, particularly the force of static friction.
Given Data
- Mass of the block (m) = 1 kg
- Coefficient of static friction (μs) = 0.50
- Acceleration due to gravity (g) = 10 m/s²
Calculating the Normal Force
The normal force (N) is affected by the vertical component of the applied force. The force can be applied at an angle (θ = 60 degrees), which introduces both horizontal and vertical components.
- Weight of the block (W) = m × g = 1 kg × 10 m/s² = 10 N
Force Analysis
When a force (F) is applied at an angle, it can be broken down into components:
- Horizontal component: F_horizontal = F * cos(θ)
- Vertical component: F_vertical = F * sin(θ)
The normal force can be expressed as:
N = W - F_vertical
N = W - F * sin(60°)
Static Friction Force
The maximum static friction force (F_friction) can be calculated as:
F_friction = μs × N
F_friction = μs × (W - F * sin(60°))
To find the force needed to overcome static friction:
F_horizontal = F_friction
Substituting values gives:
F * cos(60°) = μs × (W - F * sin(60°))
Solving for F
Substituting μs (0.50) and W (10 N):
F * 0.5 = 0.50 × (10 - F * (√3/2))
Solving this equation will yield:
F = 5.36 N
Thus, the magnitude of the force required to just start the block moving is 5.36 N, corresponding to option 'B'.
To find the force required to just start moving a block on a table, we need to consider the forces acting on the block, particularly the force of static friction.
Given Data
- Mass of the block (m) = 1 kg
- Coefficient of static friction (μs) = 0.50
- Acceleration due to gravity (g) = 10 m/s²
Calculating the Normal Force
The normal force (N) is affected by the vertical component of the applied force. The force can be applied at an angle (θ = 60 degrees), which introduces both horizontal and vertical components.
- Weight of the block (W) = m × g = 1 kg × 10 m/s² = 10 N
Force Analysis
When a force (F) is applied at an angle, it can be broken down into components:
- Horizontal component: F_horizontal = F * cos(θ)
- Vertical component: F_vertical = F * sin(θ)
The normal force can be expressed as:
N = W - F_vertical
N = W - F * sin(60°)
Static Friction Force
The maximum static friction force (F_friction) can be calculated as:
F_friction = μs × N
F_friction = μs × (W - F * sin(60°))
To find the force needed to overcome static friction:
F_horizontal = F_friction
Substituting values gives:
F * cos(60°) = μs × (W - F * sin(60°))
Solving for F
Substituting μs (0.50) and W (10 N):
F * 0.5 = 0.50 × (10 - F * (√3/2))
Solving this equation will yield:
F = 5.36 N
Thus, the magnitude of the force required to just start the block moving is 5.36 N, corresponding to option 'B'.
Free Test
FREE
| Start Free Test |
Community Answer
A block of mass 1 kg is at rest on a horizontal table. The coefficient...
R + Psin60 = Mg
R = Mg − Psin60∘
Frictional force
The body just moves when
R = Mg − Psin60∘
Frictional force
The body just moves when
 | Explore Courses for NEET exam |  |
Top Courses for NEETView all
Top Courses for NEET
Question Description
A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer?.
A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10 ms−2, then the magnitude of a force acting upwads at an angle of 60 from the horizontal that will just start the block moving is:a)5 Nb)5.36 Nc)74.6 Nd)10 NCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice NEET tests.
| Explore Courses for NEET exam | |
Top Courses for NEET
Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
