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A 10kg blovkc slide down a hill that is 200m tall with an initial speed of 12m/s. What is the final speed of the block when it reaches the bottom of the hill assuming no frictions?

Sagot :

Answer: 63.78m/s

Explanation: derive the formula of potential and kinetic converstion.

KE= ½mv² í=initial f=final g=gravity constant (9.81m/s²)

KEí+PEí=KEf+PEf

½mv²í+mgh=½mv²f+0

½v²í+gh=½v²f

V²f=v²í+2gh

√v²f=√((12m/s)²+2(9.81m/s)(2001m))

Vf=√4068

Vf= 63.78m/s