Math Problem Statement

an initially stationary block of mass [ m ] on a floor. A force of magnitude 0.500 mg is then applied at upward angle [ \theta=20.0^{\circ} ]. What is the magnitude of the acceleration of the block across the floor if (a) [ \mu_{s}=0.610 ] and [ \mu_{k}=0.510 ] and (b) [ \mu_{s}=0.410 ] and [ \mu_{k}=0.340 ]?

Solution

To determine the magnitude of the acceleration of the block across the floor, we need to analyze the forces acting on it and check if the block overcomes static friction. If it does, we then calculate its acceleration under kinetic friction.

Given Data:

  • Block mass = mm
  • Force applied F=0.500mgF = 0.500mg
  • Angle θ=20.0\theta = 20.0^{\circ}
  • Gravitational force mgmg
  • (a) Static friction coefficient μs=0.610\mu_s = 0.610 and kinetic friction coefficient μk=0.510\mu_k = 0.510
  • (b) Static friction coefficient μs=0.410\mu_s = 0.410 and kinetic friction coefficient μk=0.340\mu_k = 0.340

Step-by-Step Process:

  1. Components of the applied force:

    • Horizontal component: Fx=Fcosθ=0.500mgcosθF_x = F \cos \theta = 0.500mg \cos \theta
    • Vertical component: Fy=Fsinθ=0.500mgsinθF_y = F \sin \theta = 0.500mg \sin \theta
  2. Normal Force: The normal force NN is affected by the vertical component of the applied force: N=mgFy=mg0.500mgsinθN = mg - F_y = mg - 0.500mg \sin \theta Simplifying: N=mg(10.500sinθ)N = mg(1 - 0.500 \sin \theta)

  3. Static Friction: The maximum static friction force fsf_s is: fs=μsN=μsmg(10.500sinθ)f_s = \mu_s N = \mu_s mg(1 - 0.500 \sin \theta)

  4. Check if the block moves: The block will move if the horizontal component of the applied force FxF_x exceeds the maximum static friction force fsf_s. Compare FxF_x with fsf_s: Fx=0.500mgcosθF_x = 0.500mg \cos \theta So, if Fx>fsF_x > f_s, the block moves. Otherwise, it remains stationary.

  5. If the block moves, calculate acceleration: If the block moves, use the kinetic friction force fkf_k to find the net force and thus the acceleration: fk=μkN=μkmg(10.500sinθ)f_k = \mu_k N = \mu_k mg(1 - 0.500 \sin \theta) The net horizontal force is: Fnet=FxfkF_{\text{net}} = F_x - f_k Using Newton's second law, the acceleration aa is: a=Fnetm=Fxfkma = \frac{F_{\text{net}}}{m} = \frac{F_x - f_k}{m}

(a) For μs=0.610\mu_s = 0.610 and μk=0.510\mu_k = 0.510:

Let's calculate:

  • Fx=0.500mgcos20F_x = 0.500mg \cos 20^\circ
  • Fy=0.500mgsin20F_y = 0.500mg \sin 20^\circ
  • N=mg(10.500sin20)N = mg(1 - 0.500 \sin 20^\circ)
  • fs=μsNf_s = \mu_s N
  • fk=μkNf_k = \mu_k N

(b) For μs=0.410\mu_s = 0.410 and μk=0.340\mu_k = 0.340:

Repeat the same steps as in part (a) but with the new friction coefficients.

Let's compute the exact results.### Results:

  • (a) For μs=0.610\mu_s = 0.610 and μk=0.510\mu_k = 0.510: The block does not move, so the acceleration is 00.
  • (b) For μs=0.410\mu_s = 0.410 and μk=0.340\mu_k = 0.340: The block moves, and the acceleration is approximately 0.188m/s20.188 \, \text{m/s}^2.

Would you like further details on any of the calculations, or have any questions?

Here are 5 follow-up questions:

  1. How do different angles of applied force affect the motion?
  2. What happens if the applied force is increased?
  3. How would increasing the mass change the results?
  4. Can we calculate the work done by the applied force?
  5. How does the difference between static and kinetic friction impact acceleration?

Tip: To maximize the block's motion, minimize the angle or apply the force horizontally!

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction
Force Decomposition
Kinematics

Formulas

Horizontal component of force: Fx = F cos θ
Vertical component of force: Fy = F sin θ
Normal force: N = mg - Fy
Static friction: fs = μs N
Kinetic friction: fk = μk N
Net force: F_net = Fx - fk
Acceleration: a = F_net / m

Theorems

Newton's Second Law of Motion
Friction Theorems

Suitable Grade Level

Undergraduate Physics or Advanced High School (Grades 11-12)