A 7.93 kg box is pulled along a horizontal surface by a force F_{P} 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 box?

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

= 139.143N

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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