Demmy

## Homework Statement

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.

## The Attempt at a Solution

Mentor

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

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

Demmy

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

Mentor

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 acceleration

so 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 acceleration

so 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

Mentor

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.

Mentor

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