Problem involving spring, friction, circular motion, work, and conservation of energy

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Homework Statement


A compressed spring is released pushing a block at rest along a level, frictionless path. This is Location 1. The block then passes over a small area of friction which ends and becomes frictionless again. This is Location 2. The block then enters a loop-the-loop in the shape of a circle under no apparent acceleration. The top of this loop is Location 3. The block then completes the loop, and continues to travel on a level, frictionless path towards a 90 degree arc which launches the block vertical to a point, Location 4.

Given is the mass of the block(m), coefficient of kinetic friction([tex]\mu_k[/tex]), length of the area with friction(L), the spring constant(k), diameter of the circular(D) loop-the-loop, height of the point Location 4(y).

a) Find the work done by friction.
b) Find the distance the spring was compressed.
c) Find the normal force at the top of the loop.


Homework Equations


[tex]f_k = \mu_k m g[/tex]
[tex]W = \vec{F} \Delta \vec{r}[/tex]
[tex]K_i + U_i = K_f + U_f[/tex]
[tex]K = \frac{1}{2} m v^2 [/tex]
[tex]U = m g y[/tex]
[tex]U_s = \frac{1}{2} k x^2[/tex] - Hooke's law
[tex]\sum F_c = m a_c = \frac{m v^2}{r}[/tex]


The Attempt at a Solution



a) My current solution simply involves combining the friction force equation and putting that into the equation for work.
[tex]W_f = \mu_k m g L[/tex]

This is the energy lost from Location 1 to Location 2.

b) Using conservation of energy, find the energy required to go from Location 2 to 4.
[tex]K_2 + U_2 = K_4 + U_4[/tex]
[tex]K_2 = U_4 = m g y[/tex]

Energy at Location 0 is equal to work done by friction between Location 1 and 2 and the potential energy at height y.
[tex]K_1 = K_2 + W_f[/tex]
[tex]U_0 = K_1[/tex]
[tex]x = \sqrt{\frac{2 U_0}{k}}[/tex]

c) Converting kinetic energy to linear velocity, find the speed the block enters the loop.
[tex]r = \frac{D}{2}[/tex]
[tex]K_2 = \frac{m v^2}{2}[/tex]
[tex]v_2 = \sqrt{\frac{2 K_2}{m}}[/tex]
[tex]F_{c,top] = \frac{m v^2}{r} = m g + F_{N,top}[/tex]
[tex]F_{N,top} = \frac{m v^2}{r} - m g[/tex]
 

Answers and Replies

  • #2
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I think I needed to change my part C calculation as [tex]K_2 = K_3 +U_3[/tex]. Then solve for [tex]K_3[/tex] and then convert kinetic into velocity. So [tex]F_{N,top} = \frac{2 K_3}{r} - m g[/tex]. I think that is the correct way to calculate it.
 

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