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New Member
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Mar 27, 2009, 05:39 PM
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Using work to reduce speed
If a .5 kg mass slides across a frictionless surface at a speed of 10m/sec, how much work must be done on it to reduce its speed to 2m/sec?
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Uber Member
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Mar 27, 2009, 05:56 PM
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I would think you would need to know within how much time this needs to happen, 1 second, 20 seconds, you still have wind or is this in a vacuum?
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New Member
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Mar 27, 2009, 06:07 PM
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Umm... no that is all the information given other then the options for the answer, its multiple choice but I still have to do calculations and I don't know how to do it
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Uber Member
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Mar 27, 2009, 06:43 PM
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Don't know the answer, to slow a frictionless object, it will take braking, Or wind resistance to slow it down.
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New Member
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Mar 28, 2009, 03:05 PM
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 Originally Posted by maria_m
if a .5 kg mass slides across a frictionless surface at a speed of 10m/sec, how much work must be done on it to reduce its speed to 2m/sec?
If you calculate the kinetic energy (work) at 10m/sec and at 2m/s and subtract them the energy change is that which is required to slow it from one velocity to the other.
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New Member
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Mar 30, 2009, 11:26 AM
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work=ΔKE (change in kinetic energy)
KE(initial)= 1/2(mass)(velocity)^2= 1/2(.5)(10)^2
KE(final)= 1/2(.5)(2)^2
work= [1/2(.5)(10)^2] - [1/2(.5)(2)^2]
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