Camping equipment weighing 6000 N is pulled across a frozen lake by means of a horizontal rope. The coefficient of kinetic friction is 0.05. How much work is done by the campers in pulling the equipment 1000 m if its speed is increasing at the constant rate of 0.20 m/s²?

Question

Quizplus · Accepted Answer

The work done by the campers is equal to the force they exert on the equipment multiplied by the distance they pull it. Since the force is being exerted horizontally, only the horizontal component of the force is considered:

Fhorizontal = Fn × μk

where Fn is the normal force, which is equal to the weight of the equipment, and μk is the coefficient of kinetic friction.

Fhorizontal = 6000 N × 0.05 = 300 N

The power exerted by the campers is:

P = Fhorizontal × v = 300 N × 0.20 m/s = 60 W

The work done by the campers over a distance of 1000 m is:

W = P × t = P × d/v = 60 W × 1000 m/0.20 m/s = 3000 J

Therefore, the best choice is D, 4.2 × 10^3 J.

Camping Equipment Weighing 6000 N Is Pulled Across a Frozen