The load is moved evenly along an inclined plane 2 m long.Under the action of a force of 2.5 N directed along the plane

The load is moved evenly along an inclined plane 2 m long.Under the action of a force of 2.5 N directed along the plane, the load was lifted to a height of 0.4 m.If useful to consider that part of the work that went to increase the potential energy of the load, then the efficiency of the inclined plane in this process is 40%. What is the weight of the cargo?

The total work is equal to A = FL, the useful (spent on lifting) is 0.35 of the useful. Since the useful work is equal to the potential energy of the lifted load E = mgh, we get the equation:
0.35FL = mhg;
m = 0.35FL / (gh);
m = 0.35 * 5 * 1.6 / (10 * 0.7);
m = 0.4 kg.

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