A student decides to move a box of books into her dormitory room by pulling on a

Question

A student decides to move a box of books into
her dormitory room by pulling on a rope at-
tached to the box. She pulls with a force of
170 N at an angle of 27 above the horizon-
tal. The box has a mass of 26.4 kg, and the
coefficient of friction between box and floor is
0.258.
The acceleration of gravity is 9.8 m/s2 .
Find the acceleration of the box.
Answer in units of m/s2.

The student now starts moving the box up a
14.8 incline, keeping her 170 N force directed
at 27 above the line of the incline.
If the coefficient of friction is unchanged,
what is the new acceleration of the box?
Answer in units of m/s2.

Explanation / Answer

pulling force F = 170 N
angle = 27
mass of the box m = 26.4 kg
the coefficient of friction between box and floor is = 0.258
The acceleration of gravity is g = 9.8 m/s2
frictional force   f = [ mg - F sin ]
                          = 46.83 N
pulling force in horizontal direction f ' = F cos
                                                         = 151.47 N
net force F ' = f ' - f = 104.64 N
the acceleration of the box a = F ' / m
                                            = 3.963 m / s^ 2
(b). angle of inclination = 14.8 degrees
frictional force f = [ mg cos - F sin ]
                          = 44.62 N
pulling force f ' = F cos = 151.47 N
weight component along the incline = mg sin = 66 N
net force F ' = f ' - f - mg sin
                    = 40.85 N
the new acceleration of the box a ' = F ' / m                                                        = 1.547 m / s^ 2 pulling force F = 170 N
angle = 27
mass of the box m = 26.4 kg
the coefficient of friction between box and floor is = 0.258
The acceleration of gravity is g = 9.8 m/s2
frictional force   f = [ mg - F sin ]
                          = 46.83 N
pulling force in horizontal direction f ' = F cos
                                                         = 151.47 N
net force F ' = f ' - f = 104.64 N
the acceleration of the box a = F ' / m
                                            = 3.963 m / s^ 2
(b). angle of inclination = 14.8 degrees
frictional force f = [ mg cos - F sin ]
                          = 44.62 N
pulling force f ' = F cos = 151.47 N
weight component along the incline = mg sin = 66 N
net force F ' = f ' - f - mg sin
                    = 40.85 N
the new acceleration of the box a ' = F ' / m                                                        = 1.547 m / s^ 2                                                        = 1.547 m / s^ 2

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