A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What energy transfers and transformation occurs during the slide?b.What is the total change in the thermal energy of the slide and the seat of her pants?

Respuesta :

Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) [tex]U=668.16\ J[/tex]

Explanation:

Given:

  • mass of the child, [tex]m=24\ kg[/tex]
  • height of the slide, [tex]h=3.3\ m[/tex]
  • initial velocity of the child at the slide, [tex]v_i=0 m.s^{-1}[/tex]
  • final velocity of the child at the bottom of slide, [tex]v_f=3\ m.s^{-1}[/tex]

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

[tex]PE=m.g.h[/tex]

[tex]PE=24\times 9.8\times 3.3[/tex]

[tex]PE=776.16\ J[/tex]

Kinetic energy at the bottom of the slide:

[tex]KE=\frac{1}{2} m.v^2[/tex]

[tex]KE= 0.5\times 24\times 3^2[/tex]

[tex]KE= 108\ J[/tex]

(b)

Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.

This can be given as:

[tex]U=PE-KE[/tex]

[tex]U=776.16-108[/tex]

[tex]U=668.16\ J[/tex]