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Can someone help me with this?

### Question Description

Can you help me understand this Physics question?

A 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 box? Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service. 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 UT Austin Anonymous
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