You attach one end of a spring with a force constant k = 773 N/m to a wall and the other end to a mass m = 2.42 kg
and set the mass-spring system into oscillation on a horizontal frictionless surface as shown in the figure. To put the system into oscillation, you pull the block to a position xi = 4.96 cm
from equilibrium and release it.
(a) Determine the potential energy stored in the spring before the block is released.
(b) Determine the speed of the block as it passes through the equilibrium position.
(c) Determine the speed of the block when it is at a position xi/4.
(a) The potential energy stored in the spring before the block is released can be calculated using the equation for potential energy stored in a spring:
PE = 1/2*k*xi^2
where k is the force constant of the spring and xi is the displacement from equilibrium. Plugging in the values, we get:
PE = 1/2*773*0.0496^2 = 0.749 J
So, the potential energy stored in the spring before the block is released is 0.749 Joules.
(b) The speed of the block as it passes through the equilibrium position can be determined using the conservation of mechanical energy. At the equilibrium position, all the potential energy stored in the spring is converted to kinetic energy. Therefore,
1/2*m*v^2 = 1/2*k*xi^2
Solving for v, we get:
v = sqrt((k*xi^2)/m) = sqrt((773*0.0496^2)/2.42) = 0.719 m/s
So, the speed of the block as it passes through the equilibrium position is 0.719 m/s.
(c) The speed of the block when it is at a position xi/4 can be determined using the conservation of mechanical energy as well. At this position, the potential energy stored in the spring is converted to both kinetic energy and potential energy. The total mechanical energy at this position remains constant. Therefore,
1/2*m*v'^2 = 1/2*k*(xi/4)^2 + 1/2*m*v^2
where v' is the speed of the block at position xi/4. Plugging in the values we have calculated, we get:
v' = sqrt((k*(xi/4)^2 + m*v^2)/m) = sqrt((773*(0.0496/4)^2 + 2.42*0.719^2)/2.42) = 0.509 m/s
So, the speed of the block when it is at a position xi/4 is 0.509 m/s.