Question

If the stone loses 10% of its speed in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Answer 1

w1=200rpm*2pi/60=20.9rad/s

w2=200rpm-.10(200rpm)=180rpm

180rpm*2pi/60=18.8rad/s

a=(18.8-20.9)/10= -.21rad/s^2

I=1/2(28)(.15)^2
I=.315kg*m^2
Torque= -.21(.315)=.06615 N*m
Friction force= -.06615/.15=-.441 N
Normal force of man=.441/.2= 2.205 N
Final answer: -2.205 N