Question

A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answer 1

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:
$\sum F = ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
- the horizontal component of the force exerted by the rope, which is equal to
$F_x = F cos \theta = (250N)(\cos 32.0^{\circ} )=212.0 N$
- the frictional force, acting in the opposite direction, which is equal to
$F_f = \mu mg=(0.350)(50.0 kg)(9.81 m/s^2)=171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the box:
$F_x - F_f = ma$
$a= \frac{F_x - F_f}{m}= \frac{212.0 N-171.7 N}{50.0 kg} =0.81 m/s^2$