Question

A spring-loaded gun fires a 0.080-kg puck along a tabletop. The puck slides up a curved ramp and flies straight up into the air.

(a) If the spring is displaced 24.0 cm from equilibrium and the spring constant is 875 N/m, how high does the puck rise, neglecting friction?

x = m

(b) If instead it only rises to a height of 5.00 m because of friction, what is the change in mechanical energy?
W_nc =

Answer 1

(a) This is a question about conservation of energy. It tells you that there is no friction. So no energy is dissipated in the system. The potential energy stored in the compressed spring has to equal the gain in gravitaional potential energy at the peak height.

PE_spring=PE_grav

PE_spring=0.5*k*x^2

PE_spring=0.5*875N/m*(0.24m)^2=25.2J

PE_grav=m*g*h

PE_grav=0.09Kg*9.8m/s^2*h

h=25.2J/(0.08Kg*9.8m/s^2)=32.14 m

(b) First you have to know what mechanical energy is. Mechanical energy in a system is the sum total of kinetic and potential energy in a system. What the question is asking really is, how much energy is lost due to friction. You know the puck will rise to a height of 9.04m when energy is conserved. If the puck only rises to 5m, then the system lost mechanical energy. The loss is energy dissipated by friction, which is typically in the form of heat. Since the before and after state are both potential energy and the puck is not in mition, the mechanical energy is just the potential energy at the two states.

So you find the mechanical energy of the system before and after. The difference is the change in mechanical energy of the system.

Delta_ME=ME_after - ME_before

ME_before=0.5*k*x^2

ME_before=0.5*875N/m*(0.24m)^2=25.2J

ME_after=m*g*h

ME_after=0.08Kg*9.8m/s^2*5.00m

ME_after=3.92J

Delta_ME=-21.28 J

Delta_ME=-21.28 J

The total mechanical energy of the system is reduced by 21.28 J due to friction. This probably went into heating the puck and the surface.