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Conservation Of Energy, Block Sliding Down Incline Onto Spring.

  • #1

Homework Statement


CNf8k2D.png

A block of mass m starts from rest at a height h and slides down a frictionless plane inclined at angle θ with the horizontal, as shown below. The block strikes a spring of force constant k.
Find the distance the spring is compressed when the block momentarily stops. (Let the distance the block slides before striking the spring be l. Use the following as necessary: m, θ, k, l, and g.)

Homework Equations


Conservation of Energy:
GPE : U = mgh
K = (1/2)mv2
Spring = (1/2)kx2


The Attempt at a Solution



My Diagram:
QJ9jkix.jpg

Let H = l*sinθ
Let h = x*sinθ

Moment 1: E = mg(H+h) = mgsinθ(l+ x)

Moment 2: E = (1/2)mv2 + mgh = (1/2)mv2 + mgx*sinθ

Moment 3: E = (1/2)kx2

-------- Attempts to get rid of x2 led to
x = h/sin
(1/2)k(h/sinθ)2 = mgsinθ(l+ x) ;;;
(kh2/2sin2θ) = mgsinθ(l+ x) ;;;
(kh2/2) = mglsin3θ + mgxsin3θ ;;;
(kh2/2) - mglsin3θ = mgxsin3θ ;;;

x = k*h2/2mgsin3θ - l
 

Answers and Replies

  • #2
Doc Al
Mentor
44,912
1,170
Moment 1: E = mg(H+h) = mgsinθ(l+ x)

Moment 2: E = (1/2)mv2 + mgh = (1/2)mv2 + mgx*sinθ

Moment 3: E = (1/2)kx2
This is good.

-------- Attempts to get rid of x2 led to
x = h/sin
(1/2)k(h/sinθ)2 = mgsinθ(l+ x) ;;;
(kh2/2sin2θ) = mgsinθ(l+ x) ;;;
(kh2/2) = mglsin3θ + mgxsin3θ ;;;
(kh2/2) - mglsin3θ = mgxsin3θ ;;;

x = k*h2/2mgsin3θ - l
Don't try to express things in terms of "h"--that's an unknown directly related to "x".

Hint: Just compare moments 1 and 3. Solve that equation!
 
  • #3
ehild
Homework Helper
15,478
1,854
As h=xsinθ, your last equation does not give x explicitly in terms of the known quantities, k, m, θ, L, and g.

ehild
 
  • #4
So by solving equations 1 and 3 I got
mglsinθ+mgxsinθ = (1/2)kx2
0 = (k/2)x2 - mgsinθx - mglsinθ
quadratic formula

x = (mgsinθ ± √[mgsinθ(mgsinθ+2kl)] ) / k
 
  • #5
ehild
Homework Helper
15,478
1,854
It is all right now.

ehild
 

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