Learning Path
Question & Answer1
Understand Question2
Review Options3
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Explore TopicChoose the Best Answer
A
25 N
B
35 N
C
45 N
D
55 N
Understanding the Answer
Let's break down why this is correct
Answer
On an incline, the normal force equals the component of weight perpendicular to the surface, so \(N = mg\cos\theta\). With a mass of 10 kg, gravity is 98 N, and \(\cos30^\circ\approx0. 866\), giving \(N\approx84. 9\) N. The maximum static friction is \(\mu_s N\), so \(0.
Detailed Explanation
Normal force equals weight times the cosine of the slope angle. Other options are incorrect because This answer assumes the full weight acts as static friction, ignoring the angle of the ramp; It doubles the correct friction value by mistakenly adding the parallel weight component to the perpendicular one.
Key Concepts
Frictional Force Calculation
Static Friction Coefficient
Angle of Incline Impact
Topic
Friction Forces Analysis
Difficulty
hard level question
Cognitive Level
understand
Practice Similar Questions
Test your understanding with related questions
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Question 1When an object slides down a ramp made of rubber at a steep angle, how does the combination of the angle of incline and the material affect the frictional force experienced by the object?
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Question 2If a block weighing 20 kg is placed on a 45-degree incline, and the coefficient of static friction between the block and the incline is 0.6, what is the maximum angle at which the block can remain at rest without sliding down?
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Question 3If a 10 kg box is resting on a flat surface with a coefficient of friction of 0.5, what is the maximum frictional force that can act on the box before it starts to slide?
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Question 4In the design of a new roller coaster, engineers must consider the coefficient of friction between the coaster wheels and the tracks. If the coefficient of static friction is 0.6 and the total weight of the roller coaster is 5000 N, what is the maximum force of static friction that can be exerted before the coaster starts to slide?
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Question 5A 5 kg block is placed on a ramp inclined at an angle of 30 degrees. What is the maximum static frictional force that can act on the block if the coefficient of static friction between the block and the ramp is 0.4?
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Question 6A 15 kg box is resting on a ramp inclined at 40 degrees. If the coefficient of static friction between the box and the ramp is 0.5, what is the maximum angle of inclination at which the box can remain at rest without sliding down?
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