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

Question

Please answer using grade 11 Energy and Momentum please!

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.

Answer 1

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.