Kinetic energy and potential energy, Physics

Assumptions: Kinetic Energy=KE, Potential Energy=PE

m=mass of the skateboarder=55 kg, Velocity v=1.80m/s

So ,Initial Energy(KE) of the skateboarder=mv2/2 = 0.5* 55 * (1.8)^2 = 89.1 J

Now from law of conservation of energy we can write that

Work Done by the man/boy

=(Change in KE)+(Change in PE)+(Friction)

            =(KE final-KE initial)+(PE final- PE initial)+(Friction)

            = (KE final-KE initial)+ ?PE +(Friction)---------------------(1)                                                                                     

Now (PE final-PE initial) is to be determined.

But PE initial=0

So (PE final- PE initial)=?PE=PE final

And KE final= mv2/2=.5*55*(6)^2=990 J

KE initial=89.1 J(already determined)

So (KE final-KE initial)= 900.9 J

Work done by the man=80 J, Friction=-265 J

Now from equation (1) we get

80=900.9+?PE-265

?PE=-1085.9 J

        =-1.09KJ(approx)

This -ve sign indicates that the skateboarder losses some vertical height.

Part 2:

Change in potential energy

=?PE

=m*g*h(m=mass, g=gravitational force, h=height)

So -1.09KJ=55*9.8*h

    h= -1085.9/(55*9.8)

      =-2.02m or 2.02 m below the starting point

Posted Date: 3/20/2013 1:10:45 AM | Location : United States







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