Newtons laws, dynamics questions

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SUMMARY

This discussion focuses on applying Newton's laws of motion to solve two physics problems involving friction. The first problem involves a 1000-kg car on a 10-degree slope, where the static friction force is calculated using the equation F_x = -f_s + wsin(θ) = 0. The second problem examines a 2000-kg car skidding to a stop with a coefficient of friction of 0.4, utilizing the equation v_f^2 = v_i^2 + 2aΔx to determine the length of the skid marks. Both solutions are confirmed as correct based on the provided methodologies.

PREREQUISITES
  • Understanding of Newton's Second Law (f = ma)
  • Knowledge of static and kinetic friction concepts
  • Ability to draw and interpret free body diagrams
  • Familiarity with basic kinematic equations
NEXT STEPS
  • Study the effects of different coefficients of friction on motion
  • Learn about inclined plane dynamics and forces
  • Explore advanced kinematic equations for varying acceleration
  • Investigate real-world applications of Newton's laws in automotive safety
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of Newton's laws in action.

Nelonski
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Homework Statement


1) A 1000-kg Car is parked on a 10 degree slope. Determine the magnitude of the frictional force acting on the car

2) A 2000-kg car traveling at an initial speed of 10m/s skids to a stop with a constant acceleration. The coefficient of friction between the tired and the road is 0.4. determine the length of the skid marks


Homework Equations


Newtons 2nd law f=ma

(f_s) = static friction
f_k= kinetic friction
w= weight
n= normal force

The Attempt at a Solution



For the first part (1) . The free body diagram I drew was on a slant to show that the normal force is perpendicular to the slanted road so it is in the positive y direction, the force of static friction is to the negative x axis, and the weight force is downwards (with two components x and y)

F_x = -f_s + wsin∅ = 0
F_y= n-wcos∅=0

Essentially, I just solved the x component equation, given that wsin∅ is 9800sin10.

This was then equal to the static friction force.

This feels wrong, it was too simple, but maybe I could just be overthinking/underthinking.

Is this correct?



2)The normal force is pointing up in the positive y axis, the weight force is pointing down, and the kinetic friction force is poiting to the negative x axis(left) for the free body diagram.

Equations:

F_x= -f_k=ma_x
F_y= n-w=0

I esentially solved for the normal force in the y direction, then plugged that into f_k=μkn , to solve for the kinetic friction force. after solving the friction, I plugged that into F_x and determined the acceleration (which is negative).

After knowing the acceleration, initial and final velocities I used the equation

v_f^2=v_i^2+2aΔx

Solved for x, and think i found the answer.

Is this right as well? (the mah is entirely tough to type out so I simply gave my steps for solving)

Thanks so much
 
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Nelonski said:

Homework Statement


1) A 1000-kg Car is parked on a 10 degree slope. Determine the magnitude of the frictional force acting on the car

2) A 2000-kg car traveling at an initial speed of 10m/s skids to a stop with a constant acceleration. The coefficient of friction between the tired and the road is 0.4. determine the length of the skid marks


Homework Equations


Newtons 2nd law f=ma

(f_s) = static friction
f_k= kinetic friction
w= weight
n= normal force

The Attempt at a Solution



For the first part (1) . The free body diagram I drew was on a slant to show that the normal force is perpendicular to the slanted road so it is in the positive y direction, the force of static friction is to the negative x axis, and the weight force is downwards (with two components x and y)

F_x = -f_s + wsin∅ = 0
F_y= n-wcos∅=0

Essentially, I just solved the x component equation, given that wsin∅ is 9800sin10.

This was then equal to the static friction force.

This feels wrong, it was too simple, but maybe I could just be overthinking/underthinking.

Is this correct?



2)The normal force is pointing up in the positive y axis, the weight force is pointing down, and the kinetic friction force is poiting to the negative x axis(left) for the free body diagram.

Equations:

F_x= -f_k=ma_x
F_y= n-w=0

I esentially solved for the normal force in the y direction, then plugged that into f_k=μkn , to solve for the kinetic friction force. after solving the friction, I plugged that into F_x and determined the acceleration (which is negative).

After knowing the acceleration, initial and final velocities I used the equation

v_f^2=v_i^2+2aΔx

Solved for x, and think i found the answer.

Is this right as well? (the mah is entirely tough to type out so I simply gave my steps for solving)

Thanks so much

The method seems ok on both of those.
 

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