A child pushes a block of weight W=5.5 N against a wall with a force F acting up
ID: 2001285 • Letter: A
Question
A child pushes a block of weight W=5.5 N against a wall with a force F acting upward at =42.0 degress to the horizontal. The block is moving up the wall with a steady speed of 0.15 m/s. The coefficient of dynamic friction between the wall and the block is d=0.2. What is the magnitude of the force that the child is applying?
and
A 9.9 lb box is being pushed on a level horizontal floor with a constant velocity. The coefficient of dynamic friction between the book and the table is d= 0.2. What is the magnitude of the force of friction acting on the book?
Explanation / Answer
As the block is moving with constant speed, net force acting on block must be zero.
let F is the force applied by the child.
Apply, Fnety = 0
F*sin(beta) - W - Friction = 0
F*sin(beta) - W - F*cos(beta)*mue_d = 0
F*(sin(beta) - cos(beta)*mue_d) = W
F = W/(sin(beta) - cos(beta)*mue_d)
= 5.5/(sin(42) - cos(42)*0.2)
= 10.57 N <<<<<<--------------------------Answer
==============================================
Net force acting on the block must be zero.
Apply, Fnetx = 0
F - mue*W = 0
F = mue*W
= 0.2*9.9
= 1.98 lb <<<<<<--------------------------Answer
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