Math Problem Statement

A block with mass m = 10 kg is placed on a rough surface. A force F = 12 N is applied to the right. The coefficients of static and kinetic friction are given as μs = 0.2 and μk = 0.1, respectively. The goal is to determine: (a) the normal force (b) the frictional force (c) the acceleration of the block.

Solution

The question in the image describes a physics problem involving a block with mass m=10kgm = 10 \, \text{kg} placed on a rough surface. A force F=12NF = 12 \, \text{N} is applied to the right, and the coefficients of static and kinetic friction are provided as μs=0.2\mu_s = 0.2 and μk=0.1\mu_k = 0.1, respectively. The goal is to determine:

a) The normal force NN.

b) The frictional force between the block and the surface.

c) The acceleration of the block.

Let's go step by step to solve each part.


Given Information:

  • Mass of the block: m=10kgm = 10 \, \text{kg}
  • Applied force: F=12NF = 12 \, \text{N}
  • Coefficient of static friction: μs=0.2\mu_s = 0.2
  • Coefficient of kinetic friction: μk=0.1\mu_k = 0.1
  • Acceleration due to gravity: g=9.8m/s2g = 9.8 \, \text{m/s}^2

Part a) Normal Force NN

The normal force NN is the force exerted by the surface that balances the weight of the block. Since the block is on a horizontal surface, the normal force is equal to the weight of the block:

N=mg=10kg9.8m/s2=98NN = m \cdot g = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N}


Part b) Frictional Force

To find the frictional force, we need to determine whether the block is in motion or at rest. First, we calculate the maximum static friction force:

fstatic max=μsN=0.298N=19.6Nf_{\text{static max}} = \mu_s \cdot N = 0.2 \cdot 98 \, \text{N} = 19.6 \, \text{N}

Since the applied force F=12NF = 12 \, \text{N} is less than the maximum static friction force fstatic maxf_{\text{static max}}, the block does not move, and the static friction force exactly balances the applied force:

ffriction=F=12Nf_{\text{friction}} = F = 12 \, \text{N}


Part c) Acceleration of the Block

Since the applied force is not enough to overcome static friction, the block remains at rest. Therefore, the acceleration of the block is:

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


Summary of Results:

  • a) Normal force N=98NN = 98 \, \text{N}
  • b) Frictional force ffriction=12Nf_{\text{friction}} = 12 \, \text{N}
  • c) Acceleration a=0m/s2a = 0 \, \text{m/s}^2

Do you need further clarification or details on any part of the solution?

Here are five related questions you can explore:

  1. What would happen if the applied force FF were greater than fstatic maxf_{\text{static max}}?
  2. How do you calculate the kinetic friction force once the object starts moving?
  3. What is the relationship between normal force and frictional force?
  4. Can friction ever cause acceleration? If so, under what conditions?
  5. How would the results change if the surface were inclined instead of horizontal?

Tip: Always check if the force applied is enough to overcome static friction before assuming motion.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction
Kinematics

Formulas

Normal force: N = m * g
Friction force: f_friction = μ * N
Acceleration: F_net = m * a

Theorems

Newton's Second Law

Suitable Grade Level

High School