Question

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 83.2 N at an angle of 28.0° above the horizontal. The box has a mass of 20.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (Indicate the direction with the sign of your answer.)

(a) Find the acceleration of the box. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

magnitude	m/s2
direction	° counterclockwise from the +x-axis

(b) The student now starts moving the box up a 10.0° incline, keeping her 83.2 N force directed at 28.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? (Assume that the +x-axis is along the incline and the +y-axis is upwards and perpendicular to the incline.)

magnitude	m/s2
direction	° counterclockwise from the incline

PLEASE EXPLAIN STEPS

Answer 1

a)

Summing forces along x,

83.2 cos 28 - uk N = m a

Along y,

83.2 sin 28 - m g + N = 0

--> N = 156.94 N

Thus,

83.2 cos 28 - uk N = m a

83.2 cos 28 - 0.300 (156.94) = 20 a

Thus,

a = 1.319 m/s^2 [answer]

direction: 0 degrees counterclockwise from the +x axis [the books do not elevate!]

******************

b)

Summing forces along x,

83.2 cos 28 - uk N - m g sin 10 = m a

Along y,

83.2 sin 28 - m g cos 10 + N = 0

--> N = 153.96 N

Thus,

83.2 cos 28 - uk N - m g sin 10 = m a

83.2 cos 28 - 0.300 (153.96) - 20(9.8)sin 10= 20 a

Thus,

a = -0.338 m/s^2

However, this is contradictory, as the books are not accelerating down the incline.

THus, this means that the force is not good enough to counter friction. It is not moving. Thus,

a = 0 m/s^2

direction: 0 degrees counterclockwise from the +x axis [the books do not elevate!]

[As an alternative, if the professor overlooked this, please answer a = -0.338 m/s^2. However, please let him reconsider this problem. Thanks]