Learning Path
Question & Answer1
Understand Question2
Review Options3
Learn Explanation4
Explore TopicChoose the Best Answer
A
5 m/s
B
5 m/s
C
0 m/s
D
0 m/s
Understanding the Answer
Let's break down why this is correct
Answer
The block’s weight is 5 kg, so the normal force is 5 kg × 9. 8 m/s² ≈ 49 N, making the kinetic friction force μₖN = 0. 3 × 49 ≈ 14. 7 N. The applied push is 20 N, so the net horizontal force is 20 N – 14.
Detailed Explanation
The applied push is exactly matched by the friction that resists it. Other options are incorrect because Many think friction is small, so the block moves; Some assume the block accelerates at 1 m/s².
Key Concepts
Newton's Laws of Motion
Conservation of Momentum
Friction
Topic
Classical Mechanics Principles
Difficulty
hard level question
Cognitive Level
understand
Practice Similar Questions
Test your understanding with related questions
1
Question 1When a 15 kg block is placed on a rough surface where the coefficient of kinetic friction is 0.4, how does a significant increase in surface temperature affect the normal force and the friction acting on the block?
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Practice
2
Question 2If a 10 kg block is sliding across a horizontal surface with a coefficient of kinetic friction of 0.3, what is the kinetic frictional force acting on the block?
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Practice
3
Question 3A 5 kg block is sliding on a horizontal surface with a coefficient of kinetic friction of 0.3. If the block is pushed with a force of 20 N to the right, what is the final velocity of the block after 2 seconds, assuming it starts from rest? Consider both Newton's Laws and the conservation of momentum in your calculations.
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