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Camping equipment with a mass of 523-kg is pulled across a frozen lake by a horizontal rope. If the ice has a coefficient of kinetic friction of 0.06, how much work (in J) is done by the campers pulling their equipment 710-m at a constant velocity

Respuesta :

Answer: The work done by the campers is 218.342KJ

Explanation: Please see the attachments below

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Ver imagen Abdulazeez10
akande212

Answer:

W = 218325J of work.

Explanation:

At constant velocity the net force acting on a body is zero. The forces acting on the camping equipment are the force of friction f and pulling force F of the campers.

So F – f = 0

F = f

f = μmg

Then

F = μmg

Given

μ = 0.06, m = 523kg and g = 9.80m/s²

F = 0.06×523×9.8 = 307.5N

Workdone = F × distance = 307.5 × 710 = 218325J of work

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