We show below a free-body diagram of a 1.00-kg block pulled across a surface by

Question

We show below a free-body diagram of a 1.00-kg block pulled across a surface by a string.  The string's tension is T = 6 N.  The string makes an angle of 37 degrees with the horizontal.  The coefficient of kinetic friction is uk = 0.30.

1. Determine the y-component of the normal force.   

2. Determine the x-component of the force of friction.

3. Find the magnitude of the total force on the block.

4. Finally, find the magnitude of the acceleration of the block.

We show below a free-body diagram of a 1.00-kg block pulled across a surface by a string. The string's tension is T = 6 N. The string makes an angle of 37 degrees with the horizontal. The coefficient of kinetic friction is uk = 0.30. Determine the y-component of the normal force. Determine the x-component of the force of friction. Find the magnitude of the total force on the block. Finally, find the magnitude of the acceleration of the block.

Explanation / Answer

1) Ny=mg-T

=1*9.8-6sin37

=6.19 N

2)fx=u*N

=0.3*6.19

=1.856 N

3) F net= (Fx^2+ Fy^2)^0.5

=((4.8-1.856)^2+(9.8-3.6)^2)^0.5

=18.25 N6.86 N

4) a=F/m=fx/m

=(4.8-1.856)/1

=2.944 m/s^2

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