A metal block of mass 3 kg is moving downward with speed 2 m/s when the bottom of the block is 0.8 m above the floor. When the bottom of the block is 0.4m above the floor, it strikes the top of a relaxed vertical spring 0.4 m above the floor, it strikes the top of a relaxed vertical spring 0.4 m in length. The stiffness of the spring is 2000 N/m.

a: The block continues downward, compressing the spring. When the bottom of the block is 0.3m above the floor, what is its speed?

b: The block eventually heads back upward, loses contact with the spring, and continues upward. What is the maximum height reached by the bottom of the block above the floor?

c: What approximations did you make?

a:

Initial State: Block 0.8 m above floor, moving downward, $v_{i}$ = 2 m/s, spring relaxed

Final State: Block 0.3 m above floor, spring compressed

b:

Initial State: Block 0.8 m above floor, moving downward, $v_{i}$ = 2 m/s, spring relaxed

Final State: Block at highest point, $v_{f} = 0$, spring relaxed

a: The speed of the block when it is 0.3 m above the floor

b: The maximum height

Air resistance and dissipation in the spring are negligible.

$U_{g} \approx mgy$ near Earth's surface.

${\Delta K_{Earth}}$ is negligible.

System: Earth, block, spring

Surroundings: Nothing significant

a:

From the Energy Principle:

$E_{f} = E_{i} + W$

$K_{f} + U_{s,f} + U_{g,f} = K_{i} + U_{s,i} + U_{g,i} + W$

$U_{s,i}$ and $W$ cancel out.

$\dfrac{1}{2}mv^2_{f} + mgy_{f} + \dfrac{1}{2}k_{s}s^2_{f} = mgy_{i} + \dfrac{1}{2}mv^2_{i}$

$v_{f}^2 = v_{i}^2 + 2g(y_{i} - y_{f}) - \dfrac{k_{s}}{m}{s^2_{f}}$

$v_{f} = \sqrt {(2 m/s)^2 + 2(9.8 N/kg)(0.5 m) - \dfrac{2000 N/m}{3 kg}{(0.1 m)}^2}$

$v_{f} = 2.7 m/s$

b:

From the Energy Principle:

$E_{f} = E_{i} + W$

$K_{f} + U_{s,f} + U_{g,f} = K_{i} + U_{s,i} + U_{g,i} + W$

$K_{f}, U_{s,f}, U_{s,i}$ and $W$ cancel out.

$mgy_{f} = mgy_{i} + \dfrac{1}{2}mv^2_{i}$

$y_{f} = y_{i} + \dfrac {v_{i}^2}{2g} = (0.8 m) + \dfrac {(2 m/s)^2}{2(9.8 N/kg)} = 1.0 m$