A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the kinetic energy of the ball when it reaches 1.5 meters above the ground?

Respuesta :

Initially the total energy of the object, which is only PE.

PE = mgh = 0.2kg * 10m/s² * 3.25m = 6.5 J

When it is at 1.5m above the ground, it has both Kinetic and Potential Energy.

At 1.5m above the ground:
PE = mgh = 0.2kg * 10m/s² * 1.5m = 3 J

But at such mid level, PE + KE 

PE + KE = Total (Conserved) 

3 + KE = 6.5

KE = 6.5 - 3 = 3.5

Kinetic energy at 1.5 above ground = 3.5 J
we are given with the mass of the ball and the initial and final height of the ball when dropped from a height. The kinetic energy is obtained by first determining the final velocity of the ball. We use the formula 2ax = v^2 where a is equal to 9.8 m/s2 and x is equal to 3.25-1.5 or 1.75 meters. velocity is equal to 5.86 m/s. kinetic energy is mv2 where m is 0.20 kg. the answer is 6.87 Joules.