Question

A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of the apple at this height? 0.00 J 2.45 J 9.80 J 7.35 J

Answer 1

Answer:

2.45 JOULES

Explanation:

JUST TOOK TEST

Answer 2

The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at the bottom

$K_f$ is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

$mgh_i + K_i = mgh_f + K_f$

where:

m = 0.5 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 2 m$ is the initial height

$h_f=1.50 m$ is the final height

The initial kinetic energy is zero, since the ball starts from rest:

$K_i = 0$

Therefore we can solve the equation for $K_f$, the final kinetic energy of the ball:

$K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J$

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