This Tension in the rope must equal the weight of the supported mass, which can be easily proved using Newton’s second law... Explain this statement!! I WILL MARK BEST ANSWER AS BRAINLEST!!! PLZ NO SPAM I WILL REPORT
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Answer:
Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus,
Fnet = T − w = 0,
where T and w are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:
T = w = mg.
For a 5.00-kg mass, then (neglecting the mass of the rope) we see that
T = mg = (5.00 kg)(9.80 m/s2) = 49.0 N
If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope. Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector.
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