Friction is a repulsive force when one body is sliding through a surface. Friction is due to the contact of two surfaces. If the frictional force is strong, the motion of bodies through the surface is not smooth. Throughout this motion, the kinetic energy is converted to heat energy. The friction force is a derived force and not like gravitation i.e. a fundamental force.
Formula for friction is given as,
Ff = friction
μ = the coefficient of friction force
Fn = the normal force
It is a vector quantity.
Problem 1: An object of mass 10 kg is moving through a surface. Find the friction force if the coefficient of friction is 0.2?
Solution:
Known quantities are,
m = 10 kg and μ = 0.2
The friction force formula is,
can be calculated as,
= mg
= 10×9.81 = 98.1 N
Thus, Ff = 0.2×98.1 = 19.62 N
Friction force Ff is 19.62 N
Problem 2: Determine the friction force of a 5 kg body which moves over a surface with a friction coefficient 0.3?
Solution :
Known variables are,
m = 5 kg and μ = 0.3
The friction force formula is,
can be calculated as,
= mg
= 5×9.81
= 49.05 N
Thus, Ff = 0.3×49.05
= 14.715 N
Friction force Ff is 14.715 N
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Friction is a repulsive force when one body is sliding through a surface. Friction is due to the contact of two surfaces. If the frictional force is strong, the motion of bodies through the surface is not smooth. Throughout this motion, the kinetic energy is converted to heat energy. The friction force is a derived force and not like gravitation i.e. a fundamental force.
Formula for friction is given as,
Ff = friction
μ = the coefficient of friction force
Fn = the normal force
It is a vector quantity.
Problem 1: An object of mass 10 kg is moving through a surface. Find the friction force if the coefficient of friction is 0.2?
Solution:
Known quantities are,
m = 10 kg and μ = 0.2
The friction force formula is,
can be calculated as,
= mg
= 10×9.81 = 98.1 N
Thus, Ff = 0.2×98.1 = 19.62 N
Friction force Ff is 19.62 N
Problem 2: Determine the friction force of a 5 kg body which moves over a surface with a friction coefficient 0.3?
Solution :
Known variables are,
m = 5 kg and μ = 0.3
The friction force formula is,
can be calculated as,
= mg
= 5×9.81
= 49.05 N
Thus, Ff = 0.3×49.05
= 14.715 N
Friction force Ff is 14.715 N
Other Related Sections
NCERT Solutions | Sample Papers | CBSE SYLLABUS| Calculators | Converters | Stories For Kids | Poems for kids| Learning Concepts I Practice Worksheets I Formulas | Blogs | Parent Resource
List of Physics Formulas |
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Formula: Ptolemy’s Theorem relates the sides and diagonals of a cyclic quadrilateral. For a cyclic quadrilateral ABCD with diagonals AC and BD, the theorem states:
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