The book's initial speed = 2.83 m/s
The mass of the textbook, m = 0.75 kg
The final speed of the textbook, v = 0 m/s (Since it comes to rest)
The distance covered by the book, d = 1.2 m
Coefficient of kinetic friction, [tex]\mu = 0.34[/tex]
The frictional force between the table and the book is calculated as:
[tex]f_k=-\mu mg\\\\ f_k=-0.34 \times 0.75 \times 9.8\\\\f_k=-2.499N[/tex]
The workdone by the textbook is calculated as:
[tex]W = Fd\\\\W=2.499(1.2)\\\\W=2.9988J[/tex]
According to the work-kinetic energy theorem:
Workdone by the book = Change in Kinetic energy
[tex]W = 0.5m(v^2-u^2)\\\\-2.9988=0.5 \times 0.75 \times (0^2-u^2)\\\\-2.99988=-0.375u^2\\\\u^2=\frac{-2.9988}{-0.375} \\\\u^2=7.9968\\\\u=\sqrt{7.9968}\\\\u=2.83m/s[/tex]
The book's initial speed = 2.83 m/s
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