Question

Imagine a 4.0-kg wood block sliding across a horizontal table surface with a positive velocity and an acceleration of -2.0 m/s. You need to calculate the coefficient of kinetic friction for these two surfaces. 1. Start the same way you always should whenever you have a question involving forces and accelerations. Draw a free-body diagram (arrows for all the forces, arrow for net force, and a coordinate system) for the block. All the forces, vertical as well as horizontal. 2. The vertical component of the block's velocity is increasing, decreasing, constant but not equal to zero, constant and equal to zero). 3. Therefore, the vertical component of the block's acceleration is increasing, decreasing, constant but not equal to zero, constant and equal to zero). 4. Therefore, the vertical component of the net force on the block is increasing, decreasing, constant but not equal to zero constant and equal to zero). 5. Write the algebraic expression that shows how the forces combine to produce the net vertical force on the block. Write it as FNET vert = (algebraic expression). 6. Then set it equal to mavert. Use the numerical value of avert. After parts 2, 3, and 4, you know what that is. 7. The equation in part 6 tells you that Nl>,5 < W. 8. Write the algebraic expression that shows how the forces combine to produce the net horizontal force on the block. Write it as FNET horiz = (algebraic expression). 9. Then set that expression equal to mahoriz. Use the numerical value of a horiz. Solve that equation for the magnitude of the kinetic friction force. 10. Use your answer to part 9 and the relationship between kinetic friction and normal force to calculate the value of the coefficient of kinetic friction.

Answer 1

Solution in the uploaded image

AD Solution. N trom 4kg given figure, trun Net horizontal force. - – for – UN my Also Na mg = 40. B Fu= – 4x4x8. B 49 = mx9 = – 4 x 4x 1.8 a= - 2m/s2 O given - 2- - 4x9.8 4= 4.8 = 0.204 Caufficient of kinetic friction, 4= 0.204 AN for- MAN mg FBD (2) (3) vertical component of velocity = 0 vertical component of acceleration to From FBD; verticle component are equal and opposite Na mxg Fnet verticle =0

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