If you are pulling a block of mass 5 kg at an angle of 37 degrees and us = .55 a

Question

If you are pulling a block of mass 5 kg at an angle of 37 degrees and us = .55 and uk = .39. a) If you are pulling with a force of 10 Newtons, will the block move? If not, what is fs? b) What is the minimum force that will barely make it move?c) What is the force that will make it move at constant speed once it has moved? d) If you pull the block 1 Newton stronger than the minimum required in part b, what is its Acceleration, and the magnitude of the Normal force? e) What is the maximum you can pull the block before it gets lifted off the surface? At that point, how fast is it accelerating? f) If you pull the block 1 Newton stronger than the maximum in part e, what is its acceleration?

Explanation / Answer

pulling at an angle of 37 degrees

m=5kg , us = 0.55 , uk = 0.39

a) F = 10 N

resultant normal reaction = mg - Fsin37 = 5*9.8 - 10*sin37 = 42.98185 N

static friction = us*N = 42.98185*0.55 = 23.64 N

resultant horizontal component of force = Fcos37 = 10*cos37 = 7.98635 N

friction is more, so the block will not move

b)min force required to move

static friction = Fcos37

0.55*(mg - Fsin37) = Fcos37

F = 0.55*5*9.8/(cos37 + 0.55*Fsin37) = 23.857 N

c) constant velocity

kinetic friction = Fcos37

0.39*(mg - Fsin37) = Fcos37

F = 0.39*5*9.8/(cos37 + 0.39*Fsin37) = 18.49337 N

d) 1 N more than required

min required = 23.857 N

so F = 24.857 N

normal reaction = mg - Fsin37 = 34.0407 N

acc = [Fcos37 - 0.39*34.0407]/m = 1.315 m/s^2

e) Fsin37 = mg

F = 5*9.8/sin37 = 81.42 N

a = [Fcos37 - 0.39*(mg - Fsin37)]/m = 13.005 m/s^2

f) a = [Fcos37]/m = 13.165 m/s^2

in this case, the block is off the surface so friction does not act on block

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