How To Calculate Energy Of A Child On A Slide?

[AlphaRock]

Homework Statement

5. A 40 kg child jumps onto a playgorund slide at 3 m/s. If she slides down a vertical drop of 5.0 m while traveling 8.0 horizontally, find:

a) The speed of hte child at the bottom of the slide, if there is no friction.

b) The intial total energy of the child, relative to the bottom of the slide.

c) The amount of heat energy that must be generated if the child reaches speeds of 4.0 m/s at the bottom of the slide.

Homework Equations

Epi + Eki = Epf + Ekf
P=mgh
K=(1/2)m(v^2)

The Attempt at a Solution

a) The speed of the child at the bottom of the slide, if there is no friction.

Epi + Eki = Epf + Ekf
mgh + 0 = 0 + (1/2)m(v^2)
sqrt(2mgh/m) = v
sqrt(2gh) = v
sqrt(2*9.80*5.0) = v
sqrt(98) = v
v = 9.9 m/s?

b) The initial total energy of the child, relative to the bottom of the slide.
This one is confusing...
Epi = mgh
= 40kg * 9.80 * 5.0
= 90 J?


c) The amount of heat energy that must be generated if the child reaches speeds of 4.0 m/s at the bottom of the slide.

Heat = W = Fd = Delta Ke
Heat = Fd = (40*9.80)*(sqrt(25+64
Heat = Fd = (40*9.80)*(sqrt(89))
=400*9.4
= 4000 J?

 

[tnutty]

1) the concept looks good

2) compare the energy at the top to the bottom. Whats the difference between he two.

 

[Nabeshin]

a) The speed of the child at the bottom of the slide, if there is no friction.

Epi + Eki = Epf + Ekf
mgh + 0 = 0 + (1/2)m(v^2)
sqrt(2mgh/m) = v
sqrt(2gh) = v
sqrt(2*9.80*5.0) = v
sqrt(98) = v
v = 9.9 m/s?

Re-read the problem, Eki is not equal to zero!

b) The initial total energy of the child, relative to the bottom of the slide.
This one is confusing...
Epi = mgh
= 40kg * 9.80 * 5.0
= 90 J?

This would be the total gravitational potential energy... But the total energy is the sum of potential and kinetic as you wrote in your solution to a).

c) The amount of heat energy that must be generated if the child reaches speeds of 4.0 m/s at the bottom of the slide.

Heat = W = Fd = Delta Ke
Heat = Fd = (40*9.80)*(sqrt(25+64
Heat = Fd = (40*9.80)*(sqrt(89))
=400*9.4
= 4000 J?

No, not quite... You are sort of on the right track...
Consider the work-energy equation:
$$K_i+U_i+W=K_f+U_f$$
Where K is kinetic energy, U is potential, and W is work done on the system by an external nonconservative force.