A box weighing 40 kg slides without initial speed from the top of an inclined plane with a height of 2 m and stops. What work must be done to drag the box back along the same path to the top of the ramp?
m = 40 kg.
g = 9.8 m / s ^ 2.
h = 2 m.
When lowering the box, all of its potential energy En goes to overcome the friction force Ftr. According to the law of conservation of total mechanical energy, the work of the friction force Atr is equal to the change in the potential energy of the box.
Atr = Ep.
Potential energy is determined by the formula: En = m * g * h.
In order to raise the box to the top of the inclined plane, it is necessary to do the work to overcome the friction force Amr and to communicate the potential energy En to the box.
A = Atr + Ep = 2 * Ep.
A = 2 * m * g * h.
A = 2 * 40 kg * 9.8 m / s ^ 2 * 2 m = 1568 J.
Answer: in order to drag the box to the top of the inclined plane, you need to do work A = 1568 J.
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