How to Solve a Time-Dependent Force Problem for a Gliding Puck

AI Thread Summary
The discussion revolves around solving a time-dependent force problem involving a gliding puck with a toy rocket engine. The force exerted on the puck is given by F = (8.00 î – 4.00t ĵ) N, and the task is to determine when the puck reaches a speed of 15.0 m/s, its distance from the initial position, and the total displacement at that time. The initial calculations for acceleration components are established, with the x-component yielding 4.00 m/s² and the y-component being -2.00t m/s². Participants suggest integrating the acceleration to find velocity and combining the velocities in perpendicular directions to solve for the time, leading to a quadratic equation. The discussion highlights the complexities of combining velocities and the need for careful mathematical handling.
Leesh09
Messages
8
Reaction score
0

Homework Statement



To model a spacecraft , a toy rocket engine is securely fastened to a 2.00-kg large puck, which can glide with negligible friction over a horizontal surface, taken as the xy plane. The engine exerts a time-dependent force, F = (8.00 î – 4.00t ĵ) N, where t is in seconds, on the puck. If the puck is initially at rest, (a) At what time will it be moving with a speed of 15.0 m/s? (b) How far is the puck from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the puck traveled at this time?

Homework Equations



Fx=ma, Fy=ma, v=at

The Attempt at a Solution



we know m=2.00 kg and F=(8.00i-4.00tj)
For the x component of F, 8.00 N= (2.00 kg)(x component of acceleration)
so a sub x= 4.00 m/s^2.
For the y component: -4.00t=(2.00)(y component of a) so a sub y= -2.00t m/s^2

To find the time at which the speed is 15.0 m/s I would think you would take v=(ax+ay)t and then set v=to 15 so 15=(4.00+(-2.00t))*t so 15=4.00t-2.00t^2. This is a dead end, however, since at no point does the equation -2.00t^2+4.00t-15=0/
 
Physics news on Phys.org
Welcome to PF!

Hi Leesh09! Welcome to PF! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
Leesh09 said:
To find the time at which the speed is 15.0 m/s I would think you would take v=(ax+ay)t …

No, vx= axt, and vy= ayt,

but v is not vx + vy

how do we combine velocities in perpendicular directions? :smile:
 
would it work to say a(t)=(4.00i-2.00tj), take the integral to find velocity and then find this magnitude? so 15= sqrt(4.00t2+(-t2)2)?? so then 225=16t2+t4, getting t=3 s?
 
Leesh09 said:
… so 15= sqrt(4.00t2+(-t2)2)??

erm :redface: … where did the (t2)2 come from?
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top