Work-KE theorem

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Mango12

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Jan 17, 2016
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"Suppose the ball exits the net going straight down at 0.6m/s. What is the force of friction between the net and the ball given that the ball is in contact with the net for 0.381m? Use the Work-KE theorem. The ball is .5kg. Your answer should be close to 9N"

I know that w= change in k
and that the work done on an object by a net force equals the change in kinetic energy of the object: W = KEf - KEi

KEstart = 1/2(.5)(4.5^2)=5.06
KEend= 1/2(.5)(.6^2)=0.09

But I don't know what to do next
 
Last edited:
Mango12 said:
"Suppose the ball exits the net going straight down at 0.6m/s. What is the force of friction between the net and the ball given that the ball is in contact with the net for 0.381m? Use the Work-KE theorem. The ball is .5kg. Your answer should be close to 9N"

I know that w= change in k
and that the work done on an object by a net force equals the change in kinetic energy of the object: W = KEf - KEi

KEstart = 1/2(.5)(4.5^2)=5.06
KEend= 1/2(.5)(.6^2)=0.09

But I don't know what to do next
Click to expand...

Did you post the complete problem??!!
 
Subhotosh Khan said:
Did you post the complete problem??!!
Click to expand...

The only other info I have is that when the ball gets to the hoop it is going 4.5m/s. This is what I did so far...

1/2mv^2before = .5*.5*4.5^2=5.06
1/2mv^2after = .5*.5*.6^2=0.09

1/2mv^2before=Ffriction * S friction+1/2mv^2after
5.06= Ffriction*.381+0.09
= 4.97/.381=13

So I got an answer of 13, and the answer is supposed to be close to 9...ideas?


 
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