2 Work and Energy Problems.

jpatrie · #1 $Old$ October 8th, 2009, 04:57 PM

1. A train, of mass m, comes into a station travelling slightly too fast to stop in time before it hits the buffers at the end of the track. The buffers are effectively metal plates attached to a spring which obeys Hooke's law with a spring constant k. The trains wheels lock and it skids along the horizontal rails with sparks flying so that at the point that the train first touchs the buffers it has a speed v. If the maximum compression of the spring in the buffers is 1m what is the minimum coefficient of kinetic friction between the locked wheels of the train and the rails for the buffers to be able to stop it given that the gravitational field is g?

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2. [Difficult] Quarks are the particles that make up the proton and neutron inside an atomic nucleus. For a particular choice of units (not SI) the attractive force between two quarks is given by the equation:

where

is a constant and

is the separation between the quarks. If the potential energy, U, of the quarks is zero at equilibrium what is their potential energy as a function of

?

For this particular question, I indefinitely integrated to find:

$U = 4kr^2+k/r + C$

This is partially correct, but for some reason I cannot solve for the constant of integration. Even when I set U=0, it doesn't make sense.

Any help is greatly appreciated.

Grandad · #2 $Old$ October 9th, 2009, 06:35 AM

Hello jpatrie

Quote:

Originally Posted by jpatrie $View Post$

1. A train, of mass m, comes into a station travelling slightly too fast to stop in time before it hits the buffers at the end of the track. The buffers are effectively metal plates attached to a spring which obeys Hooke's law with a spring constant k. The trains wheels lock and it skids along the horizontal rails with sparks flying so that at the point that the train first touchs the buffers it has a speed v. If the maximum compression of the spring in the buffers is 1m what is the minimum coefficient of kinetic friction between the locked wheels of the train and the rails for the buffers to be able to stop it given that the gravitational field is g?

If the coefficient of dynamic friction is $\mu$ , then when the train is sliding, the friction force on the train is $\mu mg$ . So if this force moves through a distance $-d$ , the work it does on the train is $-\mu mgd$ .

The energy stored in a spring when it is compressed a distance $d$ is $\tfrac12kd^2$ , so the work it does on the train is $-\tfrac12kd^2$ .

The initial KE of the train is $\tfrac12mv^2$ , and when it comes to rest its final KE is 0.

So, if the train comes to rest after compressing the buffers a distance $1$ m, using the Work-Energy Principle:

$-\mu mg.1 -\tfrac12k1^2 =0 - \tfrac12mv^2$

$\Rightarrow \mu mg=\tfrac12(mv^2-k)$

$\Rightarrow$ the minimum value of $\mu$ is $\frac{mv^2-k}{2mg}$

Quote:

2. [Difficult] Quarks are the particles that make up the proton and neutron inside an atomic nucleus. For a particular choice of units (not SI) the attractive force between two quarks is given by the equation:

where

is a constant and

is the separation between the quarks. If the potential energy, U, of the quarks is zero at equilibrium what is their potential energy as a function of

?

For this particular question, I indefinitely integrated to find:

$U = 4kr^2+k/r + C$

This is partially correct, but for some reason I cannot solve for the constant of integration. Even when I set U=0, it doesn't make sense.

Any help is greatly appreciated.

The equilibrium position is when the force on the particle is zero; i.e. when

$F = 8kr - \frac{k}{r^2}= 0$

$\Rightarrow 8kr^3 = k$

$\Rightarrow r = \tfrac12$

So $U = 0$ when $r=\tfrac12$ .

$\Rightarrow 0 = 4k(\tfrac12)^2+\frac{k}{\tfrac12}+c$

$\Rightarrow c = -3k$

$\Rightarrow U =4kr^2+\frac{k}{r}-3k$

Grandad