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?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top