Spring potential energy problem

AI Thread Summary
A block with a mass of 3.5 kg is dropped from a height of 81 cm onto a spring with a spring constant of 2220 N/m, and the task is to find the maximum compression of the spring. The energy conservation principle is applied, equating gravitational potential energy to spring potential energy. The equation used is mg(h+x) = 1/2kx^2, where x represents the spring's compression. The block's initial and final kinetic energies are both zero at the points of release and maximum compression. This approach correctly accounts for the additional height due to spring displacement.
dorkymichelle
Messages
40
Reaction score
0

Homework Statement



A block of mass m = 3.5 kg is dropped from height h = 81 cm onto a spring of spring constant k = 2220 N/m (Fig. 8-37). Find the maximum distance the spring is compressed.

Homework Equations



Pe(spring) = 1/2kx2
Pe(gravity)=mgh
Ke = 1/2mv2

The Attempt at a Solution



so far, i got mgh+1/2mvi2 =1/2mvf2 = 1/2kx2

because the energy used to compress the spring is the amount of energy from the end of the drop.
but I don't have initial velocity or final velocity. or can i use 0 as final velocity at the point that the object hits the spring..
 

Attachments

  • fig08_36.gif
    fig08_36.gif
    2.4 KB · Views: 567
Physics news on Phys.org
dorkymichelle said:

Homework Statement



A block of mass m = 3.5 kg is dropped from height h = 81 cm onto a spring of spring constant k = 2220 N/m (Fig. 8-37). Find the maximum distance the spring is compressed.

Homework Equations



Pe(spring) = 1/2kx2
Pe(gravity)=mgh
Ke = 1/2mv2

The Attempt at a Solution



so far, i got mgh+1/2mvi2 =1/2mvf2 + 1/2kx2

because the energy used to compress the spring is the amount of energy from the end of the drop.
but I don't have initial velocity or final velocity. or can i use 0 as final velocity at the point that the object hits the spring..
The block is dropped from rest, and it ends up momentarily at rest when the spring is fully compressed, not when it first hits the spring. You need to change your PE term to account for the extra height due to the spring displacement in its fully compressed position.
 
so kinetic initial and kinetic final would both be 0
changing my PE(gravity) term to include the compressed spring,
would the equation I need be mg(h+x) = 1/2kx^2 ? and solve for x to be the length the spring compresses?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top