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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!