A 100 N table is pushed by a horizontal force of 40 N for 3 s. What is the velocity of the table at the end of 3 s? The coefficient of kinetic friction between the floor and table is 0.25.
Please identify the given, formula and solution..thank you!
2007-11-05 21:29:11 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
W = mg = 100N
F = 40N
t = 3s
μ = 0.25
the forces on the table are
F in the positive direction
and
friction = -μmg in the negative direction
also you know
ma = sum of the forces
so
ma = F + (-μmg)
gives
a = F/m - μg
where m = W/g so
a = Fg/W - μg = (F/W - μ)g............(*)
is the acceleration
next, you know
v = at
so substituting (*) gives
v = (F/W - μ)gt = (40/100 - 0.25)9.8*3 = 4.41m/s
,.,.,.,.,.,.,.,.,
2007-11-05 21:39:20 · answer #1 · answered by The Wolf 6 · 1⤊ 0⤋
how can the table move if the force applied is lesser than the weight of the table
2007-11-05 21:33:37 · answer #2 · answered by ^ , ^ Kloudy @_@ 2 · 0⤊ 2⤋
v = at = (F - Ff)gt/W = (F - μW)gt/W
v = (40 - 0.25*100)(9.80665)(3)/100
v â 4.4130 m/s
2007-11-05 21:59:57 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋