# Friction Equation | Problems (With Solutions)

Friction force (Ff) is equal to the product of coefficient of friction (µ) and the normal force (FN). Using the equation of friction: Ff = µ × FN, the value of friction force acting on an object can be calculated.

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Friction Practice Problems

Problem 1: A box of mass 9 kg is pushed along the floor. If the coefficient of friction between the box and floor is 0.09, then calculate the value of friction force acting on a box.

Solution:

Given data:
Mass of a box, m = 9 kg
Coefficient of friction between the box and floor, µ = 0.09
Friction force acting on a box, Ff = ?
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of friction force,
Ff = µ × FN
Ff = µ × mg (equation of normal force, FN = mg)
Ff = 0.09 × 9 × 9.81
Ff = 7.94 N

Therefore, the friction force acting on a box is 7.94 N.

Problem 2: What is the value of friction force acting on an ice slab of mass 12 kg, when it is pulled along a cement floor. The coefficient of friction between the ice slab and cement floor is 0.35.

Solution:

Given data:
Mass of an ice slab, m = 12 kg
Coefficient of friction between the ice slab and cement floor, µ = 0.35
Friction force acting on an ice slab, Ff = ?
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of friction force,
Ff = µ × FN
Ff = µ × mg (equation of normal force, FN = mg)
Ff = 0.35 × 12 × 9.81
Ff = 41.20 N

Therefore, the friction force acting on an ice slab is 41.20 N.

Problem 3: A 2 kg book resting on the desk is pushed with the hand. Calculate the friction force acting on a book, if the coefficient of friction between the book and desk is 0.24.

Solution:

Given data:
Mass of a book, m = 2 kg
Friction force acting on a book, Ff = ?
Coefficient of friction between the book and desk, µ = 0.24
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of friction force,
Ff = µ × FN
Ff = µ × mg (equation of normal force, FN = mg)
Ff = 0.24 × 2 × 9.81
Ff = 4.70 N

Therefore, the friction force acting on a book is 4.70 N.

Problem 4: Calculate the friction force acting on a 500 gm knife which is sharpened on a knife sharpening stone. The coefficient of friction between a knife and knife sharpening stone is 0.75.

Solution:

Given data:
Friction force acting on a knife, Ff = ?
Mass of a knife, m = 500 gm = 0.5 kg
Coefficient of friction between a knife and knife sharpening stone, µ = 0.75
Assume: gravitational acceleration, g = 9.81 m/s2

Using the equation of friction force,
Ff = µ × FN
Ff = µ × mg (equation of normal force, FN = mg)
Ff = 0.75 × 0.5 × 9.81
Ff = 3.67 N

Therefore, the friction force acting on a knife is 3.67 N.

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