The coefficient of kinetic friction between an 100-kg desk and the wood floor is 0.4. What force (in Newtons) must be applied to move the desk at a constant speed?

Assuming the applied force is exerted parallel to the floor, by Newton's second law both the net vertical and net horizontal forces would be zero:

∑ F (horizontal) = p - f = 0 … … … because the desk is pushed at a constant speed

∑ F (vertical) = n - mg = 0 … … … because the desk doesn't move up or down

where p is the magnitude of the applied force, f is the mag. of kinetic friction, n is the mag. of the normal force due to contact between the floor and desk, and mg is the weight of the desk. We have

n = mg = (100 kg) g = 980 N

and the mag. of friction is proportional to n according to

f = 0.4 n = 392 N

Then the applied force p has magnitude

p = f = 392 N

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hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )