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
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.