A body of mass 5 kg is placed on a rough horizontal surface. If coefficients of static and kinetic friction are 0.5 and 0.4 respectively, then find value of force of friction when external applied horizontal force is (i) 15 N (ii) 25 N and (iii) 35 N.
Answers
Answer:
Here , maximum value of static friction that can act on the body = μ
s
mg
= 25 N
(i) Since the applied force is less than the maximum possible value of static friction
therefore , the force of friction will be just enough to stop the motion of the body.
Here , friction force = 15 N
(ii) Here, applied force = maximum possible value of static friction.
Friction force = 25 N
(iii) Here , applied force > maximum value of static friction.
Therefore , the body moves and kinetic friction acts on the body.
Friction force = μ
k
mg = 20 N.
Answer:
ii) 25N
Step-by-step explanation:
Here , maximum value of static friction that can act on the body = μmg = 25 N
(i) Since the applied force is less than the maximum possible value of static friction
therefore , the force of friction will be just enough to stop the motion of the body.
Here , friction force = 15 N
(ii) Here, applied force = maximum possible value of static friction.
Friction force = 25 N
(iii) Here , applied force > maximum value of static friction.
Therefore , the body moves and kinetic friction acts on the body.
Friction force = μmg = 20 N.