October 8 Solutions

# October 8 Solutions - 7.9. IDENTIFY: SET UP: Wtot KB K A ....

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7.9. IDENTIFY: tot BA W K K . The forces on the rock are gravity, the normal force and friction. SET UP: Let 0 y at point B and let y be upward. 0.50 m A yR . The work done by friction is negative; 0.22 J f W . 0 A K . The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of the rock at this point is 2 rad / a v R , upward. EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work. (ii) 2 grav grav, grav, (0.20 kg)(9.80 m/s )(0.50 m) 0.98 J A B A W U U mgy . (b) tot grav 0 ( 0.22 J) 0.98 J 0.76 J nf W W W W . tot W K K gives 2 1 tot 2 B mv W . tot 2 2(0.76 J) 2.8 m/s 0.20 kg B W v m . (c) Gravity is constant and equal to mg . n is not constant; it is zero at A and not zero at B . Therefore, kk fn is also not constant. (d) yy F ma applied to Figure 7.9 gives rad n mg ma . 22 2 [2.8 m/s] (0.20 kg) 9.80 m/s 5.1 N 0.50 m v n m g R . EVALUATE: In the absence of friction, the speed of the rock at point B would be 2 3.1 m/s gR . As the rock slides through point B , the normal force is greater than the weight 2.0 N mg of the rock.

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## This note was uploaded on 04/03/2008 for the course PHYSICS 140 taught by Professor Evrard during the Fall '07 term at University of Michigan.

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October 8 Solutions - 7.9. IDENTIFY: SET UP: Wtot KB K A ....

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