Find magnitude of kintetic friction in following case g=10ms-2
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Normal reaction N=mg=100N
Limiting friction on the body, fL=μsN=0.5×100N=50N
(i) F=40 N is less than the limiting friction so the body is static. So ,a=0
Force of friction acting on the body is static friction, f= Driving force= 40 N
Contact force is the resultant of force of friction and normal reaction. So
C=f2+N2−−−−−−−√=(40)2+(100)2−−−−−−−−−−−−√=107.7
Step-by-step explanation:
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