Amountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. she weighs  586  n. as the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. find the tensions in the rope to the left and to the right of the mountain climber. (from the figure  α  =  67.5°  and  β  =  75.0°.) 

Given:
α = 67.5 ° , with VERTICAL, and
β = 75 °  with VERTICAL
weight of climber, W = 586 N
Need to find tensions of ropes.

Let
T1=tension to the left of climber
T2=tension to the right of climber

Solution:
A. draw FBD (free body diagram on the climber

B. using sum of horizontal forces = 0
-T1sin(alpha) +T2 sin(beta) = 0,
substitute known values
T1sin(67.5) = T2sin(75) 
0.923880T1=0.99619T2
T1=1.07827*T2.......................................(1)

C. using sum of vertical forces = 0
T1cos(alpha)+T2cos(beta)-W=0
substitute known values,
T1cos(67.5)+T2cos(75)-586=0
0.38268T1+0.087156T2-586=0..................(2)

D. Solve system of equations (1) & (2)
substitute T1 from (1) into (2)

0.38268(1.07827*T2)+0.087156T2-586=0 =>
0.49979T2-586=0
T2=1172.485
Substitute into (1)
T1=1.078273T2=1264.260

E. Check solution

Equation (1):
T1sin(alpha)=1168.024
T2sin(beta) = 1168.024   ok.

Equation (2)
T1cos(alpha)+T2cos(alpha)
=483.811+102.189
=586.000   ok

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared.