What is the total distance traveled by the mass on a compressed spring?

  • Thread starter Thread starter Jinzui
  • Start date Start date
  • Tags Tags
    Energy Springs
AI Thread Summary
The discussion revolves around calculating the total distance traveled by a 10kg mass released from a compressed spring with a spring constant of 200 N/m. The mass leaves the spring at a velocity of 3.23 m/s on a surface with a coefficient of friction of 0.15. The key equation involves balancing potential energy and kinetic energy, but participants emphasize that the spring's potential energy can be ignored after the mass is released. The correct total distance traveled by the mass, factoring in friction, is determined to be 4.35 meters. The conversation highlights the importance of focusing on the problem's details without overcomplicating the solution.
Jinzui
Messages
1
Reaction score
0

Homework Statement



A horizontal k=200N/m spring is compressed 80cm. A 10kg mass is palced at the of of the spring on a u(mew)=0.15 surface. The mass leaves the spring at the equilibrium (x=0) point with 3.23m/s. Find the total distance (starting from rest) traveled by the mass. (Answer= 4.35)

yea i got my other homework questions but this one confused me. The solution is probably simple, but i have the tendency to overthink. Thank you for the help.

Homework Equations



PEs -umg(0.8) = KE

The Attempt at a Solution



1/2(200*x2) - (0.15*10*9.8*x)= 1/2 (10*v2)

then i used the quadratic formula to find x, but i got like 0.76 i think.
 
Physics news on Phys.org
There is no need to consider the elastic potential energy; the question already tells you that the mass leaves the spring with a certain velocity, hence you can safely ignore the spring from then on.
 
Yeah, start with when the mass leaves the spring. There is no need to take the spring into consideration. It is just there to confuse you. When it comes to these problems, it's best not to over think them other wise you can get overwhelmed easily if you're just starting out. Spend time reading the problem rather than solving. Good Luck.
 
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