Please help - Motion of a hockey puck

[Demmy]

Please help -- Motion of a hockey puck

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[jedishrfu]

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).

 

[Demmy]

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

 

[Demmy]

Btw, here are the answers

a) 3.13m, 3.13m/s
b) 21.9m, 6.25m/s

 

[jedishrfu]

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

 

[Demmy]

ahm, sir, where would I get my vinitial?

 

[azizlwl]

ahm, sir, where would I get my vinitial?

Apply Newton's first law of motion.

 

[Jakeus314]

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

 

[Demmy]

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.

 

[Demmy]

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

 

[Demmy]

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.

 

[Jakeus314]

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.

try not to say you converted an acceleration into 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?

 

[Jakeus314]

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

Good.. Because no force is acting on the puck from 2s to 5s.

 

[Jakeus314]

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.

 

[Demmy]

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

 

[Demmy]

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

then I got 19.5375

 

[Jakeus314]

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

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?

 

[Jakeus314]

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

then I got 19.5375

5^2 is an error here. Has the puck been accelerating for 5 seconds?

 

[Jakeus314]

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.

 

[Jakeus314]

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.

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.

 

[Demmy]

no sir, english is my third language

 

[Demmy]

it didn't say that the the puck has stop moving, it just said that same force is applied at t=5s

 

[jedishrfu]

you have three states in this problem

1) at t0 for 2 secs puck accelerates to velocity v2 and travels dist d2

2) at t2 to t5 puck glides at constant velocity (ie v2) distance d5 = d2 + v2 * (3 secs)

3) at t5 puck accelerates to velocity v7 = v2 * a * (2 secs) and dist d7 = d5 + v2 * (2 secs) + 1/2 a * (2 secs)

 

[Demmy]

sir, I tried all of your equation I didn't got the correct answer.

and I have a question in number 2 in the equation "d5 = d2 + v2 * (3 secs)" isn't that suppose to be "a" or acceleration instead of "v" velocity? because I checked it.

 

[Jakeus314]

it didn't say that the the puck has stop moving, it just said that same force is applied at t=5s

Correct. The puck does not stop moving.

 

[Jakeus314]

sir, I tried all of your equation I didn't got the correct answer.

and I have a question in number 2 in the equation "d5 = d2 + v2 * (3 secs)" isn't that suppose to be "a" or acceleration instead of "v" velocity? because I checked it.

Their equation in #2 looks good. Using d=v*t for second piece.

 

[jedishrfu]

you have three states in this problem

1) at t0 for 2 secs puck accelerates to velocity v2 and travels dist d2

2) at t2 to t5 puck glides at constant velocity (ie v2) distance d5 = d2 + v2 * (3 secs)

3) at t5 puck accelerates to velocity v7 = v2 * a * (2 secs) and dist d7 = d5 + v2 * (2 secs) + 1/2 a * (2 secs)

slight correction on the last eqn meant to say for the last term 1/2 a *t^2 with t=2 secs