Question

A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at an angle of 25 above the horizontal. The box has a mass of 35.0kg and a the coefficient of kinetic frictino between the box and the floor is 0.27. Find the acceleration of the box

Answer 1

Answer:

$2.75 m/s^2$

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical  component of the tension in the rope (upward direction)

$mg$ is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$