the team shoots the 150g puck with an initial speed of 7.0m/s directly toward the unguarded net form a distance of 32m. The coefficient of kinetic friction for the puck on the ice is 0.08.
a) what is the force of kinetic friction acting on the puck?
b) what is the acceleration of the puck?

Question

the team shoots the 150g puck with an initial speed of 7.0m/s directly toward the unguarded net form a distance of 32m.  The coefficient of kinetic friction for the puck on the ice is 0.08.
a) what is the force of kinetic friction acting on the puck?
b) what is the acceleration of the puck?

anonymous · Answer

@experimentX ?>

ujjwal · Answer

kinetic friction is given by$$F=\mu N$$ that makes kinetic friction equal to 0.012 N ..Do you know the ans for acceleration? if its 0.76 may be i can help..

anonymous · Answer

-0.8m/s2 [toward the net] is the acceleration

anonymous · Answer

how do i calculate a and b .. im still confused

ujjwal · Answer

yeah that's a value which has been rounded off.. by using relation $$v ^{2}=u ^{2}+2as$$ v=0 in this case .. you get an acceleration of -0.765625 m/s^2 approx. 0.8 m/s^2

experimentx · Answer

looks like that's it.

experimentx · Answer

kinetic energy must be equal to work done against friction

experimentx · Answer

i got crazy result ... what is the answer for a ???

anonymous · Answer

the answer for a is 0.1N which i dont understand

experimentx · Answer

0.1148

anonymous · Answer

how did you get that #?

experimentx · Answer

thats quite in accurate.

experimentx · Answer

force * distance = ke

anonymous · Answer

the textbook is prob wrong,

experimentx · Answer

no ... the text book may be right. I am doing this type of question for the first time/

anonymous · Answer

my textbook is sometimes wrong, your probs right !

experimentx · Answer

not sure at all ... quit bg at the moment.