7.93 kg box is pulled along a horizontal surface by a force FP of 84.0 N applied at a 47.0º angle. If
the coefficient of kinetic friction is 0.35, what is the acceleration of the
AS applied force F = 84 N at an angle of 47 degrees
Then the horizontal component of force will be
F1 = 84 cos(47) = 84 * 0.682 = 57.288 N
Now normal force is
N = mg + Fsin(47)
Putting the values, we get
N = 7.93 * 9.8 + 84sin(47)
= 77.714 + 61.43
f = µN = 0.35 * 139.143 = 48.7
Now net force = F net = F1 - f = 57.288 - 48.7 = 8.588 N
F net = ma
a = F net / m
putting the values, we get
a = 8.588 / 7.93
a = 1.0829
Or a = 1.083 m/s^2
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