A 70 kg box is pulled by a 400 N at an angle of 30 degrees t

A 70 kg box is pulled by a 400 N at an angle of 30 degrees to the horizontal. the coefficient of sliding fricttion is 0.50. calculate the acceleration of the box.

Solution

Horizontal component of the applied force = 400 * cos 30° = 346 N
Normal force N = m * g + F sin 30° = m * 9.8 + 400 * 0.5 = 9.8 m + 200
Kinetic friction = µ N = 0.5 ( 9.8 m + 200 ) = 4.9 m + 100
F net = Horizontal component of the applied force - Kinetic friction
F net = 346 - ( 4.9 m + 100 )
acceleration = F net / m = [346 - ( 4.9 m + 100 )] /m

=1.47 m/s^2