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:
[tex]\sum F = ma[/tex] (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
[tex]F_x = F cos \theta = (250N)(\cos 32.0^{\circ} )=212.0 N[/tex]
- the frictional force, acting in the opposite direction, which is equal to
[tex]F_f = \mu mg=(0.350)(50.0 kg)(9.81 m/s^2)=171.7 N[/tex]
By applying Newton's law (1), we can calculate the acceleration of the box:
[tex]F_x - F_f = ma[/tex]
[tex]a= \frac{F_x - F_f}{m}= \frac{212.0 N-171.7 N}{50.0 kg} =0.81 m/s^2 [/tex]
