A 50.0 kg box is at rest on a horizontal floor whereµ k = 0.30. a) A force of 30
ID: 1722815 • Letter: A
Question
A 50.0 kg box is at rest on a horizontal floor whereµk = 0.30.a) A force of 300 N is needed to break the box free.What is the coefficient of static friction?
b) How much force is needed to move the box at aconstant speed of 4.31 m/s.
c) What would be the acceleration if a force of 500. Nis applied to the box?
a) A force of 300 N is needed to break the box free.What is the coefficient of static friction?
b) How much force is needed to move the box at aconstant speed of 4.31 m/s.
c) What would be the acceleration if a force of 500. Nis applied to the box?
Explanation / Answer
Given that the mass of box is m = 50kg coefficient of kinetic friction isµk = 0.30. ------------------------------------------------------------- (a) If the applied force for the block just slideis F = frictional force 300 N = s*mg s = 0.61 (b) If the body moving with constant velocity kinetic friction = applied force k*mg = F F = k*mg = 147 N (c) If applied force is F = 500 N then F - k*mg =ma a = ( F / m) - k*g = 7.06 m/s2
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