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## Homework Statement

A woman pushes a 38kg box on a horizontal floor with a force F⃗ that is directed 30° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.53. If the initial speed of the box is 7.89m/s and the box accelerates at 1.63m/s^2 , what is the net work done on the box by all forces acting on it after it has traveled a distance of 2.51m ?

The answer provided is 155.47J.

## Homework Equations

F=ma?

Fk = ukN

W= fdcos(x)

W= fd

W= Kf - Ki

## The Attempt at a Solution

Method 1:

F=ma

Fcos(30) - Fk = 38 (1.63)

Fcos(30) - 0.53N= 38 (1.63)

Finding the Normal Force:

N-mg-Fsin(30) =0

N =38 (9.81) + Fsin(30)

Subbing N back into the equation

Fcos(30) - 0.53[38 (9.81) + Fsin(30)]= 38 (1.63)

F = 38(1.63) + 0.53(9.81)(38) / cos(30) - 0.53sin(30)

Work: Fcos(30) x 2.51 = 1541.18J

Method 2:

W= Kf - Ki

= 0.5(38)(54.07) - 0.5(38)(62.25)

= -155.47J

Can somebody tell me why I got the wrong answer using the first method, yet when I use the second method I get the answer with a minus sign (which indicates negative work?) ?

If that is the case, then shouldn't the answer then be -155.47J?

Thank You!