Please refer to the image above to answer the question Tarza

Please refer to the image above to answer the question. Tarzan, with a mass of 60 kg, wants to swing across a ravine on a vine, but the cliff on the far side (the left side) of the ravine is 0.80 m higher than the cliff where Tarzan is now and 1.8 m higher than the lowest point in his swing. Consider Tarzan as a simple particle (no rotation and no shape changes) so that these height changes apply to his center of mass. Use g = 10 m/s2 .

(a) If Tarzan wants to reach the cliff on the far side, what is the minimum kinetic energy he must have when he leaves the cliff where he starts?

(b) Assuming Tarzan leaves the lower cliff with that minimum kinetic energy, what is Tarzan’s speed at the bottom of his swing?

(c) If Tarzan swings along a circular arc of radius 10 m, what is the tension in the vine when Tarzan reaches the lowest point in his swing? (Neglect the mass of the vine.)

(d) The work done by the force of tension acting on Tarzan as he swings is [ ] positive [ ] zero [ ] negative Briefly justify your answer:

(e) For parts (a) – (c), above, you should have assumed no air resistance acted on Tarzan. It turns out that air resistance is important here (but don’t change your answers above!), and Tarzan, when he leaves the lower cliff with the minimum kinetic energy found in (a), reaches a maximum height of just 1.2 m above his lowest point. Find the work done by air resistance during the swing for this case.

PLEASE HELP I\'M CONFUSED!! :(

Solution

a)

m = mass of tarzan = 60 kg

V_i = V_min = minimum speed on right

V_f = final speed on left = 0 m/s

h_i = initial height on right side = 0.8 m

h_f = final height on left side = 1.8 m

using conservation of energy

Total energy on right = Total energy on left

KE_i + PE_i = KE_f + PE_f

(0.5) m V_i² + mgh_i = (0.5) m V_f² + mgh_f

(0.5) V_i² + (9.8) (1) = (0.5) (0)² + (9.8) (1.8)

V_i = V_min = 3.96 m/s

b)

V_b = speed at bottom

using conservation of energy between position on right and at the bottom

Total energy on right = Total energy at bottom

KE_i + PE_i = KE_f + PE_f

(0.5) m V_i² + mgh_i = (0.5) m V_f² + mgh_f

(0.5) (3.96)² + (9.8) (1) = (0.5) V_f² + (9.8) (0)

V_f= 5.94 m/s

c)

at the lowest position forces acting are

T = tension in up

mg = weight in down

mv²/r = centripetal force in up

force equation is given as

T - mg = mv²/r

T - (60) 9.8 = 60 (5.94)² / 10

T = 799.702 N

d)

since tension force in up direction is perpendicular to the direction of motion , hence work done is 0