Math Problem Statement

Please find a simpler way to get the answer straightforward

Please find a simpler way to get the answer straightforward

Solution

Let's simplify the process step by step for filling in the blanks using the given diagram and friction equation:

Key Formulas:

  1. Friction force:
    Ffrict=μFnormF_{\text{frict}} = \mu F_{\text{norm}}

  2. Net force (Newton's Second Law):
    Fnet=ma=FappFfrictF_{\text{net}} = m \cdot a = F_{\text{app}} - F_{\text{frict}}


Steps to Solve Each Case:

Case 1: a=0m/s2a = 0 \, \text{m/s}^2

  • If acceleration a=0a = 0, the net force is zero: Fnet=0    Fapp=FfrictF_{\text{net}} = 0 \implies F_{\text{app}} = F_{\text{frict}}
  • Substitute Ffrict=μFnormF_{\text{frict}} = \mu F_{\text{norm}}: μ=FfrictFnorm\mu = \frac{F_{\text{frict}}}{F_{\text{norm}}}
  • Use the values:
    Fapp=10N,Fnorm=12NF_{\text{app}} = 10 \, \text{N}, F_{\text{norm}} = 12 \, \text{N}: Ffrict=Fapp=10NF_{\text{frict}} = F_{\text{app}} = 10 \, \text{N} μ=1012=0.833\mu = \frac{10}{12} = 0.833

Case 2: a=5m/s2(left)a = 5 \, \text{m/s}^2 \, (\text{left})

  • Acceleration is a=5m/s2a = 5 \, \text{m/s}^2 and directed left.

  • Use Fnet=maF_{\text{net}} = m \cdot a:
    Let the net force be Fnet=FfrictFappF_{\text{net}} = F_{\text{frict}} - F_{\text{app}} (since the block is accelerating left).

  • Rearrange for FfrictF_{\text{frict}}: Ffrict=Fapp+FnetF_{\text{frict}} = F_{\text{app}} + F_{\text{net}}

  • To find FnetF_{\text{net}}, calculate: F_{\text{net}} = m \cdot a = 12 \cdot 5 = 60 \, \text{N} \, (\text{mass is 12 N assuming g = 1 m/s^2}) Ffrict=16+60=76NF_{\text{frict}} = 16 + 60 = 76 \, \text{N}


Case 3: μ=0.500\mu = 0.500

  • Friction force is: Ffrict=μFnorm=0.50012=6NF_{\text{frict}} = \mu F_{\text{norm}} = 0.500 \cdot 12 = 6 \, \text{N}
  • Net force: Fnet=FappFfrict=306=24NF_{\text{net}} = F_{\text{app}} - F_{\text{frict}} = 30 - 6 = 24 \, \text{N}
  • Acceleration: a=Fnetm=2412=2m/s2a = \frac{F_{\text{net}}}{m} = \frac{24}{12} = 2 \, \text{m/s}^2

Final Answers:

  1. Case 1:

    • Ffrict=10NF_{\text{frict}} = 10 \, \text{N}
    • μ=0.833\mu = 0.833
  2. Case 2:

    • Ffrict=76NF_{\text{frict}} = 76 \, \text{N}
    • a=5m/s2(left)a = 5 \, \text{m/s}^2 \, (\text{left})
  3. Case 3:

    • Ffrict=6NF_{\text{frict}} = 6 \, \text{N}
    • a=2m/s2a = 2 \, \text{m/s}^2

Let me know if you want more detailed steps for any part!


Related Questions:

  1. How does the coefficient of friction μ\mu affect the motion of an object?
  2. What happens when Fapp<FfrictF_{\text{app}} < F_{\text{frict}}?
  3. Why is the normal force FnormF_{\text{norm}} important in friction calculations?
  4. How can we experimentally measure μ\mu in a real-world scenario?
  5. Why is the direction of acceleration opposite in Case 2?

Tip:

Always start by calculating the net force FnetF_{\text{net}}, then solve for FfrictF_{\text{frict}} and μ\mu.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Forces in Physics
Newton's Second Law of Motion
Frictional Force
Coefficient of Friction

Formulas

F_friction = μ * F_normal
F_net = m * a
F_net = F_applied - F_friction

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-12