Please Help! I only have one question! Part 3 only?

a.) What force, in newtons, must be supplied by the elevator's cable to produce an acceleration of 0.715 m/s^s upwards against a 205-N frictional force?

F = m(g + a) + Ff

F = 1250(9.81 + 0.715) + 205

F = 13361.25

F = 13.4 kN

b.) How much work, in joules, is done by the cable in lifting the elevator 19 m

If we assume the elevator car is initially at rest, work must overcome friction, change the potential energy and accelerate the mass

W = Fd + mgh + ½mv²

d = h

v² = v₀² + 2as

v² = 0² + 2(0.715)19

v² = 27.17

W = Fd + mgh + ½mv²

W = 205(19) + 1250(9.81)19 + ½(1250)27.17

W = 253863.75

W = 254 kJ

c.) What is the final speed, in meters per second, of the elevator if it starts from rest

v² = 27.17

v = 5.21 m/s

I hope this helped.

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