The 70 kg box is being pulled by a 400 N force that makes an angle of 30 degrees to the horizontal. The force of kinetic friction is 75 N. The acceleration of the box has to be determined.

The force of 400 N is applied at an angle of 30 degrees to the horizontal. The horizontal component of the force is 400*cos 30 and the vertical component is 400*sin 30 = 200

There is an opposing force due to friction that decreases the horizontal component to 400*cos 30 - 75 = 271.41 N. The acceleration of the box in the horizontal direction is 271.41/70 = 3.8773 m/s^2 and the acceleration in the vertical direction is 200/70 = 2.8571 m/s^2.

The combined acceleration has a magnitude of sqrt(3.8773^2 + 2.8571^2) = 4.8162 m/s^2 at an angle 36.38 degrees to the horizontal.

The box is accelerated at 4.8162 m/s^2 at an angle of 36.38 degrees to the horizontal.

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