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How much work must be done on a 10-kg bicycle to increase its speed from 5 ms to 10 ms?
Landish98
3000 J
Garfield Gislason
Wiki User
∙ 14y agoA. 375 j
Wiki User
∙ 15y agoThis question is all about the change in kinetic energy. Work is defined as the change in potential energy, but in this case, we can apply this to kinetic energy. Kinetic energy is defined as 1/2mv2. If we take the final kinetic energy, and subtract it from the initial kinetic energy, we will have work. In this case, 30000J-7500J=22500J Hope this helps!
Wiki User
∙ 14y ago750 j
Anonymous
3000 J
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