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?

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Physics

4 June, 21:31

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?

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Answers (1)

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Johnathan Lyons · 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