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Problem 95 Medium Difficulty

ssm A $55-\mathrm{kg}$ bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion. Find the force (magnitude and direction) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.6 $\mathrm{m} / \mathrm{s}^{2}$ Ignore the effects of air resistance.


120 $\mathrm{N}$


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Video Transcript

To solve this question, you have to use Newton's second law, and for that I would choose the following reference frame a vertical axis which I will call the Y axis. Then Newton's second law tells us that the net force that acts on the Y direction is it goes to the mass times the acceleration off the baggy jumper. But the problem tells us that the acceleration off the bank jumper is 7.6 meters per second squared downwards. So we have to include a minus sign here because off the acceleration pointing to the negative direction off our Y axis, then noticed that the net force in that direction is composed by true forces. The tension force that points to the positive direction mine is the wait for step points in the negative direction. Then the tension forced is equal to the weight force minus the mass off the bungee jumper times its acceleration the weight forces given by the mass times acceleration of gravity then detention force is given by the mass times acceleration of gravity minus the mass times acceleration and finally, detention force can be read in us M times G minus eight notice that the mass of the banking jumper According to the problem, it's 55 kilograms. Then the tension is given by 55 times, 9.8 minus 7.6 and these is 55 times 2.2 and these results in attention off approximately 120 noodles.