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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

m?

1 Answer

5 votes

Answer:

The kinetic energy is: 50[J]

Step-by-step explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.


E_(p) =m*g*h\\where:\\g= gravity[m/s^(2) ]\\m = mass [kg]\\m= (E_(p) )/(g*h)\\ m= (100)/(9.81*10)\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of


P_(e) =m*g*h\\P_(e) =1.01*9.81*5\\\\P_(e) = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

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