How Should Gravity Impact Calculations of Speed in Spring Compression Problems?

AI Thread Summary
The discussion centers on calculating the speed of a block just before it impacts a spring, emphasizing the importance of incorporating gravitational potential energy into the calculations. The initial setup incorrectly assumes that only spring potential energy is at play, neglecting the gravitational energy that contributes to the block's kinetic energy. The block's speed should be calculated by considering both the spring's compression and the gravitational potential energy lost as the block falls. The correct approach involves adjusting the energy equation to account for gravitational effects during spring compression. Understanding this relationship is crucial for accurate speed calculations in spring compression problems.
AtlBraves
Messages
11
Reaction score
0
I am having trouble with this part of the problem. I set it up like this: -.5*m*v^2 = -.5*k*d^2 so v = sqrt((k*d^2)/m) = sqrt((220*.14^2)/.25) = 4.15 m/s. What am I doing wrong?

A 250 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.2 N/cm (Figure 7-42). The block becomes attached to the spring and compresses the spring 14 cm before momentarily stopping.

(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

07_41.gif
 
Physics news on Phys.org
Possibly, you should take into account the change in gravitational potential energy as well.
 
I don't understand.
 
while the spring is compressing, gravity is still adding kinetic energy, even as the spring is taking it away. so you should see how much gravitational potential energy has been included after the 14 cm compression, and take that into account with your equation
 
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