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a 2kg object is dropped from height of 10m. ignoring air resistance calculate:

1. mechanical energy of the object

2. kinetic energy of the object when it is 3m above the ground

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Sagot :

Answer:

ME= 196.2 J

KE= 136.2

Explanation:

potential energy=mgh 2*9.81*10

Our ME is quivalent to PE as that is the toal amount of energy in the system

Kinetic energy= 1/2 m[tex]v^{2}[/tex]

to solve for kinetic enrgy we need to use a kinaetmtic equation that help us find velocity

vf= vi+at

but we need to find time first

d=vi+1/2(accelretaion)[tex]t^{2}[/tex]

7=0+1/2(9.81)[tex]t^{2}[/tex]

t= 1.19 s

vf= 0+ 9.81*1.19

vf= 11.67 m/s

Now

1/2 m[tex]v^{2}[/tex]

1/2*2*[tex]11.67^{2}[/tex]

= 136. 2

or we could just (PE/10)*7

so (196.2/10)*7

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