How much work is done ON THE BOOK witih mass of 10 kg that is being pushed across a 2m table at 5m/s?
1 answer:
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Answer with Explanation:
We are given that
a.x=-0.2 m
Where
Net electric field=E+E'=4500+1237.5=5737.5 N/C
b.x=0.20 m
Net electric field=E-E'=4500-1237.5=3262.5 N/C
c.y=0.20 m
E=1237.5 N/C
Net electric field=
F = normal force by each board on each side
W = weight of the board in between acting in down direction = 95.5 N
f = frictional force in upward direction by each board
= coefficient of friction = 0.663
Using equilibrium of force in Upward direction
f + f = W
f = W/2
f = 95.5/2 = 47.75 N
frictional force is given as
f = F
47.75 = (0.663) F
F = 72.02 N
