The speed after it has traveled 0.400 m up the wall will be 1.77 m/sec.
From the work-energy theorem, it is stated that the net work done is equal to the change in the kinetic energy.
Given data;
Mass of book,m =5.00
Force,F = 96.0 N
Θ is the angle above the horizontal = 60°
The coefficient of kinetic friction between the book and the wall is μ =0.300.
R is the normal reaction
W₁ = Fd cosΘ
W₁=(96.0 N) (0.4 m) in 60°
W₁= 33.25 J
Work done by gravity is,
W₂ = mgdcos180°
W₂ = (5.00 kg) (9.81 m/s²)(0.4 m) cos180°
W₂ = -19.6J
The Normal force on the book by the wall is,
N = FcosΘ
N = (96 N) cos 60°
N=48 J
Work done by friction is;
W₃=μkRdcos180°
W₃=0.3)(48 J)(0.4 m) cos180°
W₃ =-5.76 J
The net work done is;
W =W₁+W₂+W₃
W=33.25-19.6-5.76
W=7.89 N
From the work-energy theorem, we have
W = ΔK
[tex]\rm W = \frac{1}{2} mv^2 - 0[/tex]
The speed is;
[tex]\rm v = \sqrt \frac{2 W}{m} \\\ v = \sqrt{\frac{2 \times 7.89} {5.00}}\\\\ v = 1.77 \ m/sec[/tex]
Hence, the speed after it has traveled 0.400 m up the wall will be 1.77 m/sec.
To learn more about the work-energy theorem refer to the link;
https://brainly.com/question/16995910
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