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The net forces on the box acting perpendicular and parallel to the floor are
∑ F[perp] = F[normal] - 325 N + (475 N) sin(-35°) = 0
∑ F[perp] = (475 N) cos(-35°) - F[friction] = ma
where m is the mass of the box and a is its acceleration.
Solve for F[normal] :
F[normal] = 325 N + (475 N) sin(35°) ≈ 597 N
Then the frictional force has magnitude
F[friction] = 0.55 F[normal] ≈ 329 N
and so
60.5 N ≈ (325 N) a/g
(note that sin(-35°) = -sin(35°), cos(-35°) = cos(35°), and mg = 325 N so m = (325 N)/g)
Solve for a :
a = (60.5 N) / (325 N) g ≈ 1.82 m/s²
(a) Assuming this acceleration is constant, starting from rest, the box achieves a final velocity v such that
v² = 2a∆x
v² = 2 (1.82 m/s²) (5.80 m)
⇒ v ≈ 4.60 m/s
which happens in time t such that
v = at
4.60 m/s = (1.82 m/s²) t
⇒ t ≈ 0.177 s
(b) Let µ be the coefficient of static friction. The box just begins to slide if the magnitude of the parallel component of the applied force matches the magnitude of friction, i.e.
∑ F[para] = (475 N) cos(-35°) - F[friction] = 0
We have
F[friction] = µ F[normal] = (597 N) µ
so that
(597 N) µ = (475 N) cos(35°)
⇒ µ ≈ 0.651