What force must be applied to a body weighing 6 kg in order for it to move up an inclined plane with an acceleration

What force must be applied to a body weighing 6 kg in order for it to move up an inclined plane with an acceleration of 0.4 m / s2? The inclined plane makes an angle of 30 ° with the horizon, and the coefficient of friction is 0.3.

m = 4 t = 4000 kg.

g = 10 m / s2.

V = 2 m3.

ρw = 1000 kg / m3.

F -?

Two forces act on the stove in the water. The force of gravity m * g acts vertically downwards, buoyancy force of Archimedes Farkh vertically upwards. To lift the slab, it is necessary to apply a force F vertically upward, the value of which will be the difference between these two forces: F = m * g – Farch.

The buoyancy force of Archimedes Farch is expressed by the formula: Farch = V * ρw * g, where V is the volume of the plate, ρw is the density of water, g is the acceleration of gravity.

F = m * g – V * ρw * g = g * (m – V * ρw).

F = 10 m / s2 * (4000 kg – 2 m3 * 1000 kg / m3) = 20,000 N.

Answer: in order to lift the slab under water, it is necessary to apply a force F = 20,000 N vertically upward.