A 1,140-N crate is being pushed across a level floor at a constant speed by a fo

Question

A 1,140-N crate is being pushed across a level floor at a constant speed by a force vector F of 310 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? If the 310-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

Normal force=1140+Fsin20=1140+310*sin(20 degree)

=1246N

friction=1246*n

for constant velocity friction=Fcos20

1246*n=310*cos(20 degree)

n=0.233...............

b)normal=1140-Fsin20=1140-310*sin(20 degree)=1034N

friction=n*N=1034*0.233=240.92N

horizontal force=Fcos(20 degree)-240.92=310*cos(20 degree)-240.92=50.384N

so accelration=F/m=50.384/(1140/9.8)=0.433 m/s2 ............

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