Question

2. A student slides a 0.75 kg textbook across a table, and it comes to rest after

Answer 1

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, $\mu = 0.34$

The frictional force between the table and the book is calculated as:

$f_k=-\mu mg\\\\ f_k=-0.34 \times 0.75 \times 9.8\\\\f_k=-2.499N$

The workdone by the textbook is calculated as:

$W = Fd\\\\W=2.499(1.2)\\\\W=2.9988J$

According to the work-kinetic energy theorem:

Workdone by the book = Change in Kinetic energy

$W = 0.5m(v^2-u^2)\\\\-2.9988=0.5 \times 0.75 \times (0^2-u^2)\\\\-2.99988=-0.375u^2\\\\u^2=\frac{-2.9988}{-0.375} \\\\u^2=7.9968\\\\u=\sqrt{7.9968}\\\\u=2.83m/s$

The book's initial speed = 2.83 m/s

Learn more here: brainly.com/question/25459845