On the path of 50 m, sleds weighing 30 kg are evenly moved. Determine the work performed

On the path of 50 m, sleds weighing 30 kg are evenly moved. Determine the work performed if the friction force is 0.06 of the weight of the sled.

Given:

l = 50 meters – the distance the sled is moved;

m = 30 kilograms – sled weight;

Ftr = 0.06 * P – the ratio of the friction force to the weight of the sled;

g = 10 N / kg (Newton per kilogram) – acceleration due to gravity (rounded off value).

It is required to determine A (Joule) – the work performed on the path l.

Find the weight of the sled:

P = m * g = 30 * 10 = 300 Newtons.

Since, according to the condition of the problem, the sled is pulled evenly, the force will be equal to the friction force:

F = Ftr = 0.06 * 300 = 18 Newtons.

The work will be equal to:

A = F * h = 18 * 50 = 900 Joules.

Answer: The work done on the 50 meter section is 900 Joules.



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