A uniform .3.40 - kg rope 20.8 m long lies on the ground at the top of a vertica

Question

A uniform .3.40 - kg rope 20.8 m long lies on the ground at the top of a vertical diff. A mountain climber at the top lets down half of it to held his partner climb up the diff. What was the change in potential energy of the rope during this maneuver? A vacuum cleaner belt is looped over a shaft of radius 0.20 cm and wheel of radius 2.00 cm. The arrangement of the belt, shaft, and wheel is similar to that of the chain and sprockets in Fig. 9.15. The motor turns the shaft at 55.0 rev/s and the moving belt turns the wheel, that in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed. Assume that the belt doesn't slip on either the shaft or the wheel. What is the speed of a point of the belt? m / s What is the angular velocity of the wheel in rad/s? rad/s

Explanation / Answer

Please post question#3 separately. Here is the answer to question#2.

(2) Since only half of the rope was lowered, the hanging mass of the rope will be m = 3.40 /2 = 1.70 kg

Also of the hanging rope, the center of the mass of the rope will be acting at h = L/4 = 20.8 / 4 = 5.2 m

Change in potential energy = | 0 - mgh | = (1.70) (9.8) (5.2) = 86.632 J

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