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Answer to Question #114456 in Mechanics | Relativity for Jason Grace

Question #114456

Thank-you!

A 32.0 kg child descends a slide 4.00 m high. She reaches the bottom with a speed of 2.40 m/s. Was the mechanical energy conserved? Explain your reasoning and identify the energy transformations involved.
1
2020-05-12T09:30:26-0400

Condition: A "M =32 kg" child descends a slide "h = 4 m" high. She reaches the bottom with a speed of "v= 2.4\\frac{m}{s}". Was the mechanical energy conserved? Explain your reasoning and identify the energy transformations involved.

Solution:

By the law of conservation of energy:

"K_1+P_1=K_2+P_2", (1)

where "K_1" is a kinetic energy of child on high "h = 4 m", "K_1=0", because "v_0=0"; "P_1" is a potential energy of a child on high "h = 4 m", "P_1=Mgh"; "K_2=\\frac{1}{2}Mv^2" is a kinetic energy of a child on the bottom; "P_2=0" is a potential energy of a child on the bottom ("h'=0").

Let's rewrite (1):

"0+Mgh=\\frac{1}{2}Mv^2+0"

"Mgh=\\frac{1}{2}Mv^2" (2)

Substitute numbers from the condition to (2) and check the execution of the law (1):

"32*10*4=\\frac{1}{2}*32*2.4^2"

Divide by 32:

"10*4=\\frac{1}{2}*2.4^2"

Multiply by 2:

"10*8=2.4^2"

So, we have that

"80=5.76".

It's incorrect. The law (1) is not execute. Therefore, the part of the energy was lost for friction.

Answer: the mechanical energy is not conserved; the possible reason of it may be the friction.

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