PLease Help with the Following: A 1,380-N crate is being pushed across a level f

Question

PLease Help with the Following:

A 1,380-N crate is being pushed across a level floor at a constant speed by a force of 350 N at an angle of 20.0 degree below the horizontal, as shown in the figure (a) below. What is the coefficient of kinetic friction between the crate and the floor? micro k = If the 350-N force is instead pulling the block at an angle of 20.0 degree above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

Explanation / Answer

a) horizontal force componant = f*cos20 = 350*cos20= 328.90

normal reaction = Fsin20 + mg = 1500 N

friction force= Kinetic friction coff. * Normal reaction

and now balancing horizontal force

FCos20 = Kinetic friction coff. * Normal reaction

328.90 = kinetic friction * 1500
kinetic friction = 0.2192

2) nornam reaction = m*g-f*sin20 = 1260.30 N

now let acceleration = a

then

m*a= f*cos20 - friction coefficient* normal reaction

a= (328.90 - .2192 *1260.30 )/ ( 1380/9.8) = .3738 m/ s^2

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