How Do You Calculate the Speed of a Block Launched by Compressed Springs?

  • Thread starter Thread starter pureouchies4717
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
To calculate the speed of a block launched by compressed springs, apply the conservation of energy principle. Initially, the block has potential energy from the compressed spring, which converts to kinetic energy as it is launched. For the first scenario, use the formula for kinetic energy and potential energy to derive an expression for the initial speed v_0. In the second scenario with two springs, account for the increased spring constant, leading to a new expression for the block's speed. Comparing the two expressions will yield the relationship between v_0 and the final speed v.
pureouchies4717
Messages
98
Reaction score
0
hey guys, how would i go about solving this?

The spring in the figure (a) is compressed by length x. It launches the block across a frictionless surface with speed v_0. The two springs in the figure (b) are identical to the spring of the figure (a). They are compressed by the same length Deltax and used to launch the same block.What is the block's speed?

knight_Figure_10_38a.jpg


knight_Figure_10_38b.jpg
i know that this has to do something with v_0 raised to a power, but i can't figure out which power. please help
 
Last edited:
Physics news on Phys.org
I guess you'll need to use the conservation of energy (Etotal)before the motion= (Etotal)after the motion.

Think carefully: at the initial situation of pic A, what energy does the block contain (i.e. does it have any kinetic energy?, or potential energy?). Write down formulas for the types of energy the block has (i.e. kinetic energy= \frac{1}{2} mv^2 and add them. Now take a closer look at the variables in the formula:which values do they have in this situation?

Next consider the types of energies the block has when it has just left the spring and repeat the steps you took when you found an expression for the energy the block had at the initial situation. Be sure to equal both obtained equations and use it to obtain an expression for v_0.

Then consider situation B and do the same things you've done for situation A in order to obtain a different equation, but be sure to take into consideration that now there are two springs pushing the block, so instead of the spring constant C in part A you now are dealing with 2C. The equation you get from analyzing situation B can be written down in a way so you get an expression for v.

Now you only have to take a look at the expression of v_0 and compare it to the one you have for v.

Good luck! And if you don't understand some of my explanation, don't hesitate and please ask!

EDIT: :P Just found out you've already solved the problem ;)
 
Last edited:
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