Question

A 33-kg child jumps to the ground from the top of a fence that is 1.7 m high. You analyze the problem using upward as the positive x direction. Part A Taking x = 0 to be at the bottom of the fence, what are the initial potential energy of the child-Earth system and the change in the system kinetic energy during the jump? Enter your answers numerically separated by a comma.

Answer 1

U = 550 J and Em1 = ​​551

Step-by-step explanation:

In the child's jump, the mechanical energy is conserved, which we will write the energy in several points. Let's start with the mechanical energy Enel highest and lowest point

Initial, highest point

Emo = U = mg h

Let's calculate

Emo = U = 33 9.8 1.7

U = 550 J

These are the potential energy at the highest point, at this point the kinetics is zero as the child descends the potential energy decreases and the kinetics increases

Final lowest point

Em1 = ​​K = ½ m v²

With energy it is conserved

Em₀ = Em₁

m g h = ½ m v²

Here we can calculate the speed of the body

v = √ 2gh

let's calculate

v = √ 2 9.8 1.7

v = 5.78 m / s

Another point could be some jump point

Em₃ = K + U = ½ m v₃² + mgy₃

mg h = ½ m v₃² + m g y₃

V₃² = 2 g (h-y₃)

Let's calculate a specific point when it is halfway up and 3 = h / 2

V₃² = Ra 2 g h / 2

v₃ = Ra 2 9.8 1.7 / 2

v₃ = 4.08 m / s

Let's calculate the kinetic energy at the lowest point

Em1 = ​​K = ½ 33 5.78 2

Em1 = ​​551 J

We can see that it is the same within the approximations