A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force Fvec of magnitude 6.3 N and a vertical force Pvec are then applied to the block. The coefficients of friction for the block and surface are ?s = 0.44 and MU k = 0.20. (a) Determine the magnitude of the frictional force acting on the block if the magnitude of P vector is 8.0 N. ____ N (b) Determine the magnitude of the frictional force acting on the block if the magnitude of P vector is 10.0 N. ____ N (c) Determine the magnitude of the frictional force acting on the block if the magnitude of P vector is 12.0 N. ____ N

Explanation / Answer

( a )

N - mg + P = 0 ;

N = mg - P ;

N = 2.5 * 9.8 - 8 = 16.5 ;

fs = 0.44 * 16.5 = 7.26 N ;

fk = 0.2 * 16.5 = 3.3 N ;

since applied force is less than fs , so the

friction force = 6.3 N <-----------ans

( b )

N = mg - P ;

N = 2.5 * 9.8 - 10 = 14.5 N ;

fs = 0.44 * 14.5 = 6.38 ;

fk = 0.2 * 14.5 = 2.9 N

since applied force = 6.3 < fs ( = 6.38 N ) ;

so friction force = fs = 6.3 N <-------------ans

( c )

N = mg - P ;

N = 2.5 * 9.8 - 12 = 12.5 ;

fs = 0.44 * 12.5 = 5.5 N ;

fk = 0.2 * 12.5 = 2.5 N ;

since 6.3 > fs ( =5.5 N ) ,

so fk will act

friction force = 2.5 N <----------------ans

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