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masicld
Feb 26, 2007, 03:59 PM
:confused: :confused: A force of 30.0 N is required to start a 4.4 kg box moving across a horizontal concrete floor.
(a) What is the coefficient of static friction between the box and the floor?

(b) If the 30.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?

Capuchin
Feb 27, 2007, 12:39 AM
a) nma = F, where n is the coefficient of friction

b) same here, but you need to work out the force being lost to friction.

(use F=ma with a and m, then using nF1 = F2 work out the friction)

Mandeep Kumawat
Jul 1, 2007, 06:06 AM
a.) F=nMG
0r 30=n X 4.4 X 9.8
or Static friction coff(n)=30/(4.4X9.8)

b.) F=kMG+MA
30=K(4.4X9.8)+(4.4X0.5)

or coff of kinetic friction=(30-2.2)/(4.4X9.8)