A load weighing 10 kg moves on a horizontal surface under the action of a force of 100 N directed at an angle

A load weighing 10 kg moves on a horizontal surface under the action of a force of 100 N directed at an angle of 3000 to the horizon. The friction coefficient is 0.1. Determine the acceleration of the load.

The force that acts on the body is calculated by the formula F = m * a – Ftr – m * g – N. Expand on coordinate axes

ОX: F * cos A – K * N = m * a

OY: F * sin A + N = m * g, N = m * g – F * sin A

Then the acceleration is found by the formula a = (F * cos A – k * (m * g – F * sin A)) / m = (100 * √3 / 2 – 0.1 * (10 * 10 -100 * 1/2) ) / 10 = 8.1 m

Answer: acceleration is 8.1 m / s ^ 2.