How Far Will the Spring Compress When a Block is Dropped?

jedjj
Messages
26
Reaction score
0

Homework Statement


A block of mass m = 2.0 kg is dropped from height h = 55 cm onto a spring of spring constant k = 1960 N/m. Find the maximum distance the spring is compressed.

Homework Equations


[tex]\Delta K+ \Delta U_G+ \Delta U_S=0[/tex] \change in kinetic energy+change in gravitational energy+change in spring energy=0
[tex]\Delta K=0[/tex]
[tex]U_{Gf}-U_{Gi}+U_{Sf}-U_{Si}=0[/tex]
[tex]U_{Si}=0[/tex] can be assumed
so
[tex]U_{Sf}=U_{Gi}-U_{Gf}[/tex]

The Attempt at a Solution


resumed from above:
[tex]\frac {kx_f^2}{2}=mgy_i-mgy_f[/tex]
so: [tex]\frac {kx_f^2}{2}=mgy_i-mgx_f[/tex]
With this I have tried over and over to solve for [tex]x_f[/tex] but I cannot find any way of getting [tex]x_f[/tex] by itself.
edit:LaTeX formatting
 
Physics news on Phys.org
conservation of energy, [tex]\Delta[/tex]U(g)+ [tex]\Delta[/tex]U(spring) = 0 <=== look, all the energy lost from one object goes to the other, so the sum is 0. Remember! conservation of energy. Think about it.

[tex]\Delta[/tex]U(g)=-[tex]\Delta[/tex]U(spring) <===== moved that equation around a it... just algebra.

mgh2-mgh1 = (1/2)kx^2 <===== look what we have here if we make it look more detailed with what we know.

Or simply mgh=-(1/2)kx^2
 
Last edited:
[tex]\frac {kx_f^2}{2}=mgy_i-mgx_f[/tex]
With this I have tried over and over to solve for [tex]x_f[/tex] but I cannot find any way of getting [tex]x_f[/tex] by itself.
It's a quadratic equation.
 

Similar threads

Replies
7
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
4K
Replies
29
Views
3K
Replies
9
Views
3K