Reply Fri 26 Sep, 2014 01:08 pm
A 5.7-kg steel ball is dropped from a height of 25m into a box of sand and sinks 0.20m into the sand before stopping. How much energy is dissipated through the interaction with the sand?
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Type: Question • Score: 1 • Views: 7,246 • Replies: 2
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engineer
 
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Reply Fri 26 Sep, 2014 02:29 pm
@volleyballgurl48,
The ball has potential energy when it is in the air. That energy is converted to kinetic energy (speed) while it is falling and then dissipated into the sand when it lands. Since energy is conserved, you only need to know the potential energy it had at the beginning of the problem.

P = mgh = (5.7 kg)(9.8 m/s^2)(25m) = 1397 kg m^2/sec^2 = 1397 J
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futureastrophysicist
 
  1  
Reply Sun 7 Apr, 2024 10:59 pm
@volleyballgurl48,
potential energy = mgh
m = mass of steel ball (5.7-kg)
g = acceleration due to gravity (gravitation constant) (9.8-m/s^2)
h = 25-m + 0.20-m (include the 0.20-m because the indent in the sandbox that the steel ball made needs to be accounted for)

therefore: mgh = (5.7-kg)(9.8-m/s^2)(25-m + 0.20-m)
= (5.7-kg)(9.8-m/s^2)(25.20-m)
= 1407.672J
≈ 1407J or 1410J
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