QQuestionPhysics
QuestionPhysics
A 15.0 kg box is released on a 32 -degree incline and accelerates down the incline at 0.30 m/s².
Find the friction force impeding its motion.
What is the coefficient of kinetic friction?
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Answer
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Step 1:I'll solve this problem step by step using the precise LaTeX formatting guidelines:
Step 2:: Identify the given information
- Acceleration due to gravity ($$g$$) = 9.8 m/s²
- Angle of incline (\theta) = 32°
Step 3:: Calculate the components of gravitational force
- Perpendicular to incline: $$F_{g\perp} = mg \cos(\theta)
The gravitational force components are:
Step 4:: Apply Newton's Second Law along the incline
\sum F = ma
mg \sin(\theta) - f = ma
Step 5:: Calculate the normal force
N = mg \cos(\theta)
Step 6:: Use friction force equation
f = \mu_{k} N
Step 7:: Solve for friction force
f = (15.0 \text{ kg})(9.8 \text{ m/s}^{2}) \sin(32°) - (15.0 \text{ kg})(0.30 \text{ m/s}^{2})
Rearranging the Newton's Second Law equation: Substituting known values:
Step 8:: Calculate friction force
f = 73.2 \text{ N}
f = 77.7 - 4.5
Step 9:: Calculate coefficient of kinetic friction
\mu_{k} = 0.33
\mu_{k} = \frac{73.2}{(15.0 \times 9.8 \times \cos(32°))}
Final Answer
- Friction Force: 73.2 N - Coefficient of Kinetic Friction: 0.33
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