Question icon
Grade 7Mechanics

A grinder machine whose wheel has a radius of 7/22 is rotating at 2.5rev/sec. a tool to be sharped is held against the wheel with a force of 40N. If the coefficient of the friction between the tool and the wheel is 0.2, power required is?

Profile image of Pranee
7 Years agoGrade 7
Answers icon

1 Answer

Profile image of Kalpak
7 Years ago
W (omega ) =2πf
f=2.5 rev/ sec
v= RW
Now,  v= 7/22 * 2 * 22/7 * 2.5 (R= 7/22 given )
               =5 SI unit 
Normal force (N) =40N
Frictional force (f)  = u N
                                   = 0.2 * 40 
                                    = 8 N
Therefore , Power = f v
                                 = 8 * 5 =40 J