Velocity released from a spring

  • Thread starter Thread starter PhizKid
  • Start date Start date
  • Tags Tags
    Spring Velocity
AI Thread Summary
When a block is released from a compressed spring on a frictionless surface, the velocity at the moment it loses contact with the spring can be determined using energy principles. The work done by the spring is negative during compression and positive when the spring decompresses, leading to a change in kinetic energy. The equation used is -(1/2)kx^2 = (1/2)m(v_f)^2, where k is the spring constant and x is the compression distance. The confusion arises from interpreting the sign of the work done by the spring, which is negative when compressing but positive when the block is released. Understanding that the energy stored in the spring is positive helps clarify the calculations.
PhizKid
Messages
477
Reaction score
2

Homework Statement


Everything is on a frictionless horizontal surface. If a block is compressed against a spring for some distance x meters and then released, what is the velocity of the block the moment it loses contact with the spring?


Homework Equations


W_s = -(1/2)kx^2 [if reference point is 0m]
W = (1/2)m(v_f)^2 - (1/2)m(v_i)^2


The Attempt at a Solution


So if we take the compression of the spring to be our 0 point, then when it is released, the block travels x meters when it is fully released by the spring. The work done by the spring on the block is going to be negative. But since the initial velocity is 0, we get the equation: -(1/2)kx^2 = (1/2)m(v_f)^2. If you try to solve this for the velocity, you have to take the square root of a negative number. What am I doing wrong here?
 
Physics news on Phys.org
PhizKid said:
So if we take the compression of the spring to be our 0 point, then when it is released, the block travels x meters when it is fully released by the spring. The work done by the spring on the block is going to be negative.
Why do you think that the work done by the spring will be negative?

Which way does the spring push? Which way does the block move?
 
The displacement is in the same direction as the force of the spring pushing on it, so it should be positive, but then why does the textbook say that the work done by a spring is always -(1/2)kx^2 if the starting position is taken to be the 0 point?
 
PhizKid said:
The displacement is in the same direction as the force of the spring pushing on it, so it should be positive, but then why does the textbook say that the work done by a spring is always -(1/2)kx^2 if the starting position is taken to be the 0 point?
The work done by the spring is negative when you compress it from 0 to x and positive when it decompresses from x to 0 (which is what is going on here).

I prefer to think in terms of the energy stored in the spring, which is simply 1/2kx^2.
 
It says that the work done is negative because the block or the system DOES work ON the spring AGAINST its restoring force.
 
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