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?
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.