A 265 kg load is lifted 18.0 meters vertically by a single cable with an acceler

Question

A 265 kg load is lifted 18.0 meters vertically by a single cable with an acceleration of a = 0.160 * g (where g is the acceleration due to gravity g = 9.8 m/s/s) From Newton's 2nd Law: What is the Tension in the cable ? (draw a free body diagram and write Newtons 2^nd Law) What is the work done by the cable? (due to the Tension found in part a) (Total energy change) What is the work done by gravity on the load? (Change in potential energy) If the load starts at zero velocity, what is the velocity of the load after it is raised 18 meters? There are at least two different ways to solve this part.

Explanation / Answer

a) Let tension in cable is FT N.

FT - mg = ma

FT = m*(g+a)

FT = 3012.52 N

b) Work is done by cable on the load = FT*18 = 3012.52*18 = 54225.36 J

c) Work is done by Gravity on the load = -m*g*18 = -46746 J

d) Method-1:- by work energy theorem

total work is done by all forces on load = change in kinetic energy of load

54225.36 - 46746 = m*v2/2

v2 = 56.448

v = 7.51 m/s

Method-2:- by using 3rd equation of kinematics.

2*a*h = vf2 - vi2

vi = 0 (given)

let vf is v.

2*0.16*g*18 = v2

v2 = 56.448

v = 7.51 m/s

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