Experiment 6: Static and Kinetic Friction Wooden block Push Figure 2 Draw a free

Question

Experiment 6: Static and Kinetic Friction Wooden block Push Figure 2 Draw a free body diagram for the block as it slides and slows down. From that diagram and Newton's second law, derive an expression for the magnitude of the kinetic frictional force f in terms of the block mass m and the block's acceleration a. Also, use the free-body diagram to find the normal force; use that and the relation between fi and n to derive an expression for the coefficient of kinetic friction k n terms of a and the acceleration of gravity g. Write those expressions in the equation table below. 2 fs

Explanation / Answer

applied force = F
mass of block = m
so for moving distance x, energy applied = Fx
now, if acceleration of the block is a, then velocity gained from rest = v
KE gained = 0.5mv^2
now energy applied = KE gained + frictional KE
frictional KE = Fx - 0.5mv^2
but 2*a*x = v2
hence
Friciontal KE = Fx - m*ax = fk*x [ where fk is kinetic friction]
hence
fk = F - ma

then , weight of the object = mg
from newtons third law
normal reaction on the block due to floor = N = mg
now friction force on the block, f = k*mg ( where k is coefficient of friction)

from neewtons second law
F - f = m*a ( where a is acceleration of the body)
F - k*mg = ma
k = (F - ma)/mg

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