Friday, April 27, 2012

A skier of mass 72.3 kg is pulled up a slope by a motor-driven cable.How much work is required to pull the skier 61.9 m up a 37.0 degree slope...

When any load is pulled up a slope the total effort required and the work done has two components - to overcome the the frictional resistance and to move load against force of gravity.


In this case the slope is assumed to be frictionless. therefore the force of friction and work done against that is 0. The only force against which the skier is pulled up is the force of gravity.


This problem can have two solution based on the way the data on distance of 61.9 m is interpreted. It could be either (1) vertical distance moved, or (2) distance along the slope over which the skier is pulled up. We will solve the problem for both these assumptions.


The other given data is:


Angle of slope = A = 37 degrees.


Mass of skier = m = 73.2 kg


Acceleration due to gravity = g = 9.81 m/s^2


Constant speed = 2.0 m/s


Solution when 61.9 meter is the vertical height:


The work done = m*g*vertical distance moved


= 73.2*9.81*69.1 = 49620.1572 J


Solution when 61.9 meter is the length of slope:


The vertical distance moved = s


= (Length of slope)*Sin A = 61.9*0.6018 = 41.58438 m


The work done = m*g*s = 29861.4106 J


This is work dine neglecting the work done to accelerate the skier to speed of 2 meters per second. If we take this in to consideration the kinetic energy of skier wen being pulled up is given by


Kinetic energy = 1/2*m*v^2 = (1/2)*73.2*(2^2) = 146.4 J


Considering kinetic energy of speed of skier:


Total work done = 29861.4106 + 146.4 = 33967.8 J

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