Please help -- Motion of a hockey puck 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 x-axis; 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
Re: Please help 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).
Re: Please help 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
Re: Please help -- Motion of a hockey puck 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
Re: Please help -- Motion of a hockey puck 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.
Re: Please help -- Motion of a hockey puck @Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir
Re: Please help -- Motion of a hockey puck 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.
try not to say you converted an acceleration in to a velocity... You may find a velocity using an acceleration etc. V = V0 + a*t V from pt A is good for V0 for pt B why?
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.
Re: Please help -- Motion of a hockey puck ahm sir, sorry bout the convertion thingy, I don't get is sir.
Re: Please help -- Motion of a hockey puck 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
Maybe I'm just being picky. When people say it that way I start thinking they're missing some algebra/units/equation concepts. Is English not your first language?
Check my post 14 again. Do pt B by adding three things. Pt A answer, d=r*t during the 3 seconds of no acceleration, and last add the distance traveled while accelerating for 2 seconds with a starting velocity of ptA velocity.
I can't say for sure why you got acceleration units. Do unit analysis to see what it should be, given the values you added or multiplied.