How High Does a Stone Reach When Shot from a Slingshot?

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SUMMARY

This discussion focuses on the physics of projectile motion and energy conservation, specifically analyzing a stone shot from a slingshot and various scenarios involving potential and kinetic energy. The calculations confirm that a 12g stone reaches a height of 1.8m at the top of its arc when shot at 36m/s, and a 27kg child climbing a 5.0m slide has a potential energy of 1323J. Additionally, the maximum speed of a roller coaster descending from a height of 94.5m is calculated to be 43m/s. The discussion also highlights the importance of accounting for kinetic energy at different points in motion.

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  • Understanding of basic physics concepts such as kinetic energy (Ek) and potential energy (Eg).
  • Familiarity with the equations of motion and energy conservation principles.
  • Knowledge of projectile motion and the effects of gravity on objects in motion.
  • Ability to perform calculations involving mass, velocity, and height in physics problems.
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  • Investigate the effects of friction on motion and energy loss in mechanical systems.
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Aka
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i have a test tomorow and I am having trouble with these questions can some one pls check over my answers?
1. a slingshop shoots a 12g stone at 36m/s. It goes in an arc and at the top of the arc it is going at 14m/s. How high is it at this point?
12g=0.012kg
36m/s=v1
14m/s=v2
Ek=1/2mv2
=1/2(0.012kg)(36m/s)2
=0.216J
Eg=mgh
0.216=0.012kg(9.8m/s)h
0.216J
0.1176
=1.8m
the height is 1.8m
2. A child of 27kg climbs to the top of a 5.0m high playground slid
A) what is her potentian energy?
27kg=m
5.0m=h
Eg=mgh
=27kg(9.8m/s)(5.0m)
=1323J
the potential energy is 1323J
B)Her actual speed at the bottom is only 3m/s. According to conservation of energy how fast should se be going?
Ek=1/2mv2
=1/2(27kg)v2
1323/13.5=v2
98=v2
9.9=v
she should be going 9.9m/s
C) the slid is 8.0m long. determine the force of friction acing
d=8.0m
Ff=?
d=(v2+v1/2)t
t=1.6s
a=v2-v1/t
a=9.9m/s-0m/s/1.6s
a=6.2m/s2
F=ma
27kg(6.2m/s2)
F=167.4N
the force of friction is 167.4N
3. A 72kg passenger is in a van moving at 15m/s when it colides with a wall. The frount part of the van collapses 0.50m during the crash. determine the work done on the passenger and the force the seatbelt exerts during the acident.
72kg=m
15m/s=v1
0.50m=d
w=?
Ff=?
Ek=1/2mv2
=1/2(72kg)(15m/s)2
=8100J
...now what?
4/ A roler coaster has a height of 94.5m above the ground at its highest point. deteremine the maximum speed of the cars as they come down from this high point. the actual speed is 41.1m/s. Calcuate the work done by friction.
Eg=mgh
=1kg(9.8m/s2)(94.5m)
=926.1J
Ek=1/2mv2
926.1=1/2(1kg)v2
926.1/0.5=v2
1852.2=v2
43m/s=v
the maximun speed it 43m/s
i don't know how to calculate the force of friction
 
Last edited:
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For the first question you forgot to take into account the fact that the projectile still has kinetic energy at the top so [itex]PE_{top} = KE_{bottom} - KE_{top}[/tex][/itex]
 

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