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 force is only slightly larger than the frictional force, giving a small positive net force. Other options are incorrect because This answer assumes the block moves fast enough to reach 5 m/s, ignoring the friction that slows it; Like option A, it treats the applied force as if friction did not matter.
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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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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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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