8.1 Potential energy of a system  (Page 6/9)

Strategy

Solution

1. Since the gravitational potential energy is zero at the lowest point, the change in gravitational potential energy is
   
   $\text{Δ}{U}_{\text{grav}}=mgy-0=(12\text{N})(5.0\text{cm})=0.60\text{J}.$
2. The equilibrium position of the spring is defined as zero potential energy. Therefore, the change in elastic potential energy is
   
   $\text{Δ}{U}_{\text{elastic}}=0-\frac{1}{2}k{y}_{\text{pull}}^2=\text{−}\left(\frac{1}{2}\right)\left(6.0\frac{\text{N}}{\text{m}}\right){(5.0\text{cm})}^2=\mathrm{-0.75}\text{J}.$
3. The block started off being pulled downward with a relative potential energy of $0.75\text{J}.$ The gravitational potential energy required to rise $5.0\text{cm is}0.60\text{J}$ . The energy remaining at this equilibrium position must be kinetic energy. We can solve for this gain in kinetic energy from [link] ,
   
   $\text{Δ}K=\text{−}(\text{Δ}{U}_{\text{elastic}}+\text{Δ}{U}_{\text{grav}})=\text{−}(\mathrm{-0.75}\text{J}+0.60\text{J})=0.15\text{J}.$

Significance

A sample chart of a variety of energies is shown in [link] to give you an idea about typical energy values associated with certain events. Some of these are calculated using kinetic energy, whereas others are calculated by using quantities found in a form of potential energy that may not have been discussed at this point.