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The net forces on the box acting parallel and perpendicular to the incline, respectively, are
∑ F[para] = mg sin(14.0°) - F[friction] = ma
∑ F[perp] = F[normal] - mg cos(14.0°) = 0
where m = mass of the box and a = its acceleration.
The second equation tells us
F[normal] = mg cos(14.0°)
so that friction has a magnitude of
F[friction] = 0.380 F[normal] = 0.380 mg cos(14.0°)
Solve the first equation for a :
mg sin(14.0°) - 0.380 mg cos(14.0°) = ma
a = g sin(14.0°) - 0.380 g cos(14.0°)
a ≈ -1.24 m/s²
Assuming the box slides down the incline with constant acceleration, solve for ∆x, the distance it slides before coming to rest:
0² - (1.70 m/s)² = 2a ∆x
∆x = -(1.70 m/s)² / (2 (-1.24 m/s²))
∆x ≈ 1.16 m