An acrobat of mass m slides down a vertical rope of height h. for the first three quarter of her descent she grips the rope with her hands and legs so as to produce a fictional force equal to five-ninth of her weight. she then tighten her grip so that she comes to rest at the bottom of the rope. sketch a (t .v) graph to illustrate her descent

Using the work-energy theorem would be best in solving this problem. The net work on an object causes a change in the kinetic energy of the object, based from the work-energy theorem. For this, the formula for net work = change in kinetic energy = final kinetic energy - initial kinetic energy.

So, first sliding distance covered 3/4 with a net force 1-5/9 = 4/9 and w (w stands for weight of acrobat) 

Therefore, work done = 4/9 * 3/4
In the next step, the distance to be covered is 1/4. 
So, let the new frictional force required be X w 
Therefore, net force upward = (X - 1) w 
Thus, work to be done = (X - 1) * 1/4 w(2) 
Equating X = 1 + 4/3 = 7/3 

So, the frictional force that she must produce in the last quarter has to be 7/3 times her weight.

answer: sine

step-by-step explanation: