A 20 kg metal block is placed on a horizontal
surface. The block just begins to slide when
horizontal force of 100 N is applied to it. Calculate
the coefficient of static friction. If coefficient of
kinetic friction is 0.4 then find minimum force to
maintain its uniform motion.
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Answer:-
As we know than
According to Newton's second low of motion:-
Given:-
m = 20 kg
As we know than
According to Newton's second low of motion:-
Given:-
m = 20 kg
Solution:-
As ⇒ F = μs * mg
⇒ 10 = μs * 20 * 9.8
⇒ μs = 10 / 20*9.8
⇒ μs = 0.51
The coefficient of status friction is μs = 0.51
As ⇒ F = μs * mg
⇒ 10 = μs * 20 * 9.8
⇒ μs = 10 / 20*9.8
⇒ μs = 0.51
The coefficient of status friction is μs = 0.51
According to Newton's second low of motion:-
F = ma
As it is given that acceleration = 0
So,
⇒ F - kinetic energy = m * a
⇒ F - Kinetic energy = 0
⇒ F - μk * mg = 0
⇒ F - 0.4 * 20 * 9.8 = 0
⇒ F - 78.4 = 0
⇒ F = 78.4 N.
F = ma
As it is given that acceleration = 0
So,
⇒ F - kinetic energy = m * a
⇒ F - Kinetic energy = 0
⇒ F - μk * mg = 0
⇒ F - 0.4 * 20 * 9.8 = 0
⇒ F - 78.4 = 0
⇒ F = 78.4 N.
So, The coefficient of static friction if coefficient of K.F is 0.4 is 78.4 N
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