Question

1) You place a book of mass 5.00 kg against a vertical wall. You apply a constant force F⃗ to the book, where F = 99.0 N and the force is at angle of 60.0∘ above the horizontal (Figure 1) . The coefficient of kinetic friction between the book and the wall is 0.300.

a) If the book is initially at rest, what is its speed after it has traveled 0.400 m up the wall.

Answer 1

Here ,

mass of book, m = 5 Kg

F = 99 N

theta = 60 degree

coefficient of kinetic friction, u = 0.300

a)

Normal force acting on the book

N = F * cos(60)

N = 99 * cos(60)

N = 49.5 N

let the final speed of the book is v m/s

Using Work energy theorum

0.5 * m * v^2 = F * sin(theta) * d - m*g * d - u * N * d

0.5 * 5 * v^2 = 99* sin(60) * 0.4 - 5 * 9.8 * 0.4 - 0.300 * 49.5 * 0.4

solving for u

u = 1.87 m/s

the speed of the book after travelling 0.4 m is 1.87 m/s