A 525-g book is placed on a wooden plank which is at an angle of 35.0° to the horizontal. After the book is released, it is measured to slide down the plank with an acceleration of 2.56 m/s2.

(a) Calculate the coefficient of kinetic friction between the book and the plank.
(b) If the book was released from rest, calculate the work done by gravity after it has moved down the plank a distance of 55.0 cm.
(c) If the book was released from rest, calculate the work done by friction after it has moved down the plank a distance of 55.0 cm.

m = 0.525 kg

weight = w = m*g Newtons
normal force = m g cos 35
weight component down ramp =m g sin 35
friction up ramp = mu * m g cos 35
m a = m * 2.56 = m g sin 35 - mu m g cos 35
mu = ( g sin 35 - 2.56)/ (g cos 35)

b) m g *change in altitude = m g * 0.55 sin 35

c) mu * m g cos 35 * 0.55

To solve this problem, we can start by analyzing the forces acting on the book while it is sliding down the plank.

(a) The forces acting on the book are the force of gravity (mg), the normal force (N), and the kinetic friction force (f_k):

- The force of gravity (mg) acts vertically downwards and has a magnitude of (525 g) * (9.8 m/s^2) = 5145 N.
- The normal force (N) acts perpendicular to the plank and opposes the force of gravity. Since the plank is at an angle to the horizontal, the normal force can be found using the equation N = mg * cos(theta), where theta is the angle between the plank and the horizontal. Therefore, N = (525 g) * (9.8 m/s^2) * cos(35.0°).
- The kinetic friction force (f_k) acts parallel to the plank and opposes the motion of the book. We can find the magnitude of the kinetic friction force using the equation f_k = μ_k * N, where μ_k is the coefficient of kinetic friction. Therefore, f_k = μ_k * [(525 g) * (9.8 m/s^2) * cos(35.0°)].

In this problem, the acceleration of the book down the plank is given as 2.56 m/s^2. Since the net force acting on the book is equal to the product of its mass and acceleration (F_net = ma), we have:

F_net = (525 g) * (9.8 m/s^2) - μ_k * [(525 g) * (9.8 m/s^2) * cos(35.0°)] = (525 g) * (2.56 m/s^2).

Solving for μ_k, we can find the coefficient of kinetic friction.

(b) To calculate the work done by gravity after the book has moved down the plank a distance of 55.0 cm (or 0.55 m), we can use the work-energy theorem. The work done by gravity is equal to the change in the book's gravitational potential energy, which can be calculated as:

Work_gravity = m * g * h,

where h is the vertical distance the book has moved down the plank. In this case, h is equal to the product of the distance moved down the plank (0.55 m) and the sine of the angle between the plank and the horizontal (sin(35.0°)).

Work_gravity = (525 g) * (9.8 m/s^2) * 0.55 m * sin(35.0°).

(c) To calculate the work done by friction after the book has moved down the plank a distance of 55.0 cm (or 0.55 m), we can use the definition of work done by a constant force: Work_friction = f_k * d, where d is the distance moved down the plank.

Work_friction = μ_k * [(525 g) * (9.8 m/s^2) * cos(35.0°)] * 0.55 m.

Now we can use these formulas to find the coefficient of kinetic friction (a), the work done by gravity (b), and the work done by friction (c).