1. A 30.0-kg child starting from rest slides down a water slide with a vertical
ID: 1789227 • Letter: 1
Question
1. A 30.0-kg child starting from rest slides down a water slide with a vertical height of 20.0m. If friction is ignored, a) find the child's speed halfway down the slide's vertical distance and (b) find the speed 15 m down the slides' vertical distance (c) find the speed at bottom the slides' vertical distance? (d) If the speed down at the bottom is only 5m/s and the energy lost is due to friction (friction is not ignored), find the energy lost (for d, we haven't given any example yet, see if you can do i)Explanation / Answer
This question is based on conservation of energy.
This means that the total energy of the system remains the same.
Total Energy= Potential Energy+ Kinetic energy
TE= PE+ KE
PE= mass*g*height= mgh
KE= mass*(velocity)2 /2 = mv2/2
Keeping this equation in mind, let us approach the initial situation.
The child is at rest when at the top of the verticle slide, therefore speed is zero and hence KE=0
TE=PE= 30 kg* 9.8 * 20 m = 5880 J
a) When the child is halfway down the verticle slide, the height is 10 m
As the child has travelled some distance, the potential energy has been converted to kinetic energy.
However, the total energy will remain the same.
5880= 30*9.8*10 + KE
KE= mv2/2 = 2940 J
2940 J = 30 * v2 /2
2940/15 = v2
196= v2
v=14 m/s
b) 15 m down the slide's verticle distance i.e. h=15 m
the total energy will remain the same.
5880= 30*9.8*15 + KE
KE= mv2/2 = 1470 J
1470 J = 30 * v2 /2
1470/15 = v2
98= v2
v=9.89 m/s
c) When the child reaches the bottom of the slides, entire potential energy is converted to kinectic enerry.
TE= KE
5880 J= 30 * v2 /2
5880/15= v2
392=v2
v= 19.78 m/s
d) If the speed down at the bottom is 5m/s, the corresponding KE= 30*52/2 = 375 J.
Therefore, the difference between this KE and the KE from part c) is the energy lost due to friction as the total energy has to remain constant.
Energy lost due to friction = 5880-375= 5505J
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