Examples

We should think about using energy (conservation or work-energy) whenever the answer requested for a given problem is independent of time. The reason for this is clear: we derived the work-energy theorem (and energy conservation) from the elimination of $t$ from the dynamical equations.

1. Suppose we have a block of mass $m$ being pulled by a string at a constant tension $T$ at an angle $\theta$ with the horizontal along a rough table with coefficients of friction $\mu_s > \mu_k$. Typical questions might be: At what value of the tension does the block begin to move? If it is pulled with exactly that tension, how fast is it moving after it is pulled a horizontal distance $L$?
   

   We see that in the $y$-direction, $N + T \sin(\theta) - mg = 0$, or $N = mg - T \sin(\theta)$. In the $x$-direction, $F_x = T \cos(\theta) - \mu_s N = 0$ (at the point where block barely begins to move).

   Therefore:
   

   $$T \cos(\theta) - \mu_s mg + T \mu_s \sin(\theta) = 0$$ (3.13)

   
   
   or
   

   $$T = \frac{\mu_s mg}{\cos(\theta) + \mu_s \sin(\theta)}$$ (3.14)

   
   

   With this value of the tension T, the work energy theorem becomes:
   

   $$W = \Delta E_k$$ (3.15)

   
   
   
   

   $$\left(T \cos(\theta) - \mu_k(mg - T \sin(\theta)\right) L = \frac{1}{2}m v_f^2 - 0 \quad (v_i = 0)$$ (3.16)

   
   
   or
   

   $$v_f = \left(\frac{2 \mu_s g L \cos(\theta)}{\cos(\theta) + \... ...heta)}{\cos(\theta) + \mu_s \sin(\theta)}\right)^\frac{1}{2}$$ (3.17)

   
   

   Although it is difficult to check exactly, we can see that if $\mu_k = \mu_s$, $v_f = 0$ (or the mass doesn't accelerate). This is consistent with our value of $T$ - the value at which the mass will exactly not move against $\mu_s$ alone.

2. We can also combine energy conservation with Newton's laws for different parts of the problem:

   Suppose we have a spring gun with a bullet of mass $m$ compressing a spring with force constant $k$ a distance $\Delta x$. When the trigger is pulled, the bullet is released from rest. It passes down a horizontal, frictionless barrel and comes out a distance $H$ above the ground. What is the range of the gun?

   

   Part I: We find the speed of the bullet leaving the gun barrel from work-energy:
   

   $$W = \int_{x_1}^{x_0} -k(x - x_0)dx$$ (3.18)

   $$= -\frac{1}{2}k(x - x_0)^2 \vert _{x_1}^{x_0}$$ (3.19)

   $$= \frac{1}{2}k (\Delta x)^2 = \frac{1}{2}m v_f^2 - 0$$ (3.20)

   
   
   or
   

   $$v_f = \sqrt{\frac{k}{m}}\vert\Delta x\vert$$ (3.21)

   
   

   Part II: Given this speed (in the $x$-direction), find the range from Newton's Laws:
   

   $$\vec{F} = -m g \hat{y}$$ (3.22)

   
   
   or $a_x = 0$, $a_y = -g$, $v_{0x} = v_f$, $v_{0y} = 0$, $x_0 = 0$, $y_0 = H$. Solving as usual, we find:
   

   $$R = v_{x0} t_0$$ (3.23)

   $$= v_f \sqrt{\frac{2H}{g}}$$ (3.24)

   $$= \sqrt{\frac{2kH}{mg}} \vert\Delta x\vert$$ (3.25)