M*g = 0.525 * 9.8 = 5.15 N. = Wt. of book.
Fp = 5.15 * sin35 = 2.95 N. = Force in-parallel with plank.
Fn = 5.15*Cos35 = 4.21 N. = Normal force or force perpendicular to plank.
a. Fp - Fk = M*a.
2.95 - Fk = 0.525 * 2.56,
Fk = 1.61 N. = Force of kinetic friction.
Fk = u*Fn = 1.61.
u * 4.21 = 1.61,
u = ?
b. W = Mg * d = 0.525 * 0.55 =
c. W = Fk * d =
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.
1 answer