Question

You place a book of mass 5.00 kg against a vertical wall. You apply a constant force F⃗ to the book, where F = 96.0 N and the force is at angle of 60.0∘ above the horizontal. The coefficient of kinetic friction between the book and the wall is 0.300.
If the book is initially at rest, what is its speed after it has traveled 0.400 m up the wall.

Answer 1

The speed after it has traveled 0.400 m up the wall will be 1.77 m/sec.

What is the work-energy theorem?

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

$\rm W = \frac{1}{2} mv^2 - 0$

The speed is;

$\rm v = \sqrt \frac{2 W}{m} \\\ v = \sqrt{\frac{2 \times 7.89} {5.00}}\\\\ v = 1.77 \ m/sec$

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;

brainly.com/question/16995910

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