
#1
Aug2812, 11:41 AM

P: 12

1. The problem statement, all variables and given/known data
Hi guys, can you please help me answer this problem? A hockey puck with mass 0.160 kg is at rest at the origin on the horizontal, frictionless surface of the rink. At time t=0 a player applies a force of 0.250 N to the puck, parallel to the xaxis; he continues to apply this force until t=2.00 s. (a) What are the position and speed of the puck at t=2.00 s? (b) If the same force is again applied at t=5.00 s, what are the position and speed of the puck at t=7.00s I got the 1st one, but I couldn't get the second one. 2. Relevant equations 3. The attempt at a solution 



#2
Aug2812, 11:49 AM

P: 2,493

welcome to PF!
You need to show your work before we can help. So why not show us how you solved for (a) and where you got stuck for (b). 



#3
Aug2812, 11:53 AM

P: 12

Given: F=0.250 N
m=0.160 kg F=ma a=F/m leads to: a=0.250/0.160 i got a= 1.5625 then I used the formula a=(2s)/(t^2) to get the distance then I got s=3.13 m then, couldn't solve for b, please help 



#4
Aug2812, 12:17 PM

P: 12

Please help  Motion of a hockey puck
Btw, here are the answers
a) 3.13m, 3.13m/s b) 21.9m, 6.25m/s 



#5
Aug2812, 10:10 PM

P: 2,493

extend (a) to 5 secs what is dist and velocity then these become the initial conditions in your eqns.
so at 2 secs the force ends and the puck glides along at a constant velocity (assume no friction) for 3 secs (so you're now at 5 sec mark) vfinal = vinitial + a * t where t =7  5 dfinal= dinitial + vinitial * t + 1/2 a * t 



#6
Aug2812, 11:11 PM

P: 12

ahm, sir, where would I get my vinitial?




#7
Aug2812, 11:22 PM

P: 961





#8
Aug2812, 11:25 PM

P: 48

Should V initial at 5s be any different than V at 2s? Can you find V at 2s?




#9
Aug2812, 11:30 PM

P: 12

sir, I got the final velocity
I converted a= 1.563 m/(s^2) to m/s so I got v= 3.13 m/s then I used it as my vinitial to get my vfinal using the equation you gave me, my vfinal= 6.256 m/s but when I tried to substitute that to the 2nd equation you gave me, I didn't got the correct answer. 



#10
Aug2812, 11:31 PM

P: 12

@Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir




#11
Aug2812, 11:37 PM

P: 12

I have another question, I got letter a) but why is the unit I got is m/(s^2) the answer should be m/s, but I got the right answer. so I just randomly thought about converting the acceleration to velocity, because the unit of acceleration is m/(s^2) and the velocity is m/s, but if I do that in a) the unit would be correct but the answer would be wrong.




#12
Aug2812, 11:37 PM

P: 48

V = V0 + a*t V from pt A is good for V0 for pt B why? 



#13
Aug2812, 11:38 PM

P: 48





#14
Aug2812, 11:42 PM

P: 48

The position in pt B comes from three distances traveled while moving. Pt A distance contributes some, then some more while moving at a constant speed, then more while accelerating again. Find the position for pt B from those pieces and you'll do fine.




#15
Aug2812, 11:49 PM

P: 12

ahm sir, sorry bout the convertion thingy, I don't get is sir.




#16
Aug2812, 11:53 PM

P: 12

I tried to find the distance for 5s, so I used the same acceleration
so my equation is (1.563*(5^2))/2 then I got 19.5375 



#17
Aug2812, 11:54 PM

P: 48

Is English not your first language? 



#18
Aug2812, 11:57 PM

P: 48




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