PHYSICS HELP A worker at a loading dock tries to save steps by pushing three cra
ID: 1577711 • Letter: P
Question
PHYSICS HELP
A worker at a loading dock tries to save steps by pushing three crates at once. The crates stand side by side, and the worker pushes on the crate closest to him. The crates have masses of 40 kg, 35 kg, and 25 kg. The coefficient of kinetic friction between the crates and the ground is 0.5. (a) How much force must exert on the crates to give them an acceleration of 0.4 m/s2 ? (b) How much force does the first crate exert on the second crate? (c) How much force does the second crate exert on the third crate?
Solve two ways. Acceleration goin to the right and then to the left.
Explanation / Answer
in vertical,
N = m g = (40 + 35 + 25)(9.8)
N = 981 N
f = uk N = 490.5 N
Fnet = F - f = m a
F - 490.5 = (40 + 35 + 25)(0.4)
F = 530.5 N ...........Ans (it will be same for both cases.)
(B) for acc. to the right :
F - F12 - f = m a
530.5 - F12 - (0.5 x 40 x 9.81) = 40 x 0.4
F12 = 318.3 N .....Ans
for acc. to be to the left:
F12 - f = m a
F12 - (0.5 x 40 x9.81) = 40 x 0.4
F12 = 212.2 N .....ANs
(C) to the right:
F23 - (0.5 x 25 x 9.8) = 25 x 0.4
F23 = 132.6 N
to the left:
530.5 - F23 - (0.5 x 25 x 9.8) = 25 x 0.4
F23 = 398 N
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