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--- 183_notes:examples:a_rebounding_block [2014/10/28 06:14] – pwirving
+++ 183_notes:examples:a_rebounding_block [2014/10/31 21:39] (current) – pwirving
@@ Line 1: / Line 1: @@
-===== Example: Thermal Equilibrium - Part A =====
+===== Example: A Rebounding Block =====
-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 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 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?
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 === Facts ===
+Metal block of mass 3 kg
+The stiffness of the spring is 2000 N/m.
+The block has the following initial conditions for both parts:
 a:
@@ Line 48: / Line 54: @@
 Surroundings: Nothing significant
+Energy Principle:  $E_{f} = E_{i} + W$
+ $K =  \dfrac{1}{2}mv^2$
+ $U_{spring} = \dfrac{1}{2}k_{s}s^2$
+ $U_{gravitational} = mgy$
+{{183_notes:examples:mi3e_07-019.jpg?200|}}
 === Solution ===
@@ Line 56: / Line 71: @@
  $E_{f} = E_{i} + W$
+The total final energy is the sum of the kinetic energy, spring potential energy and gravitational potential energy which is equal to the total initial energy which is the sum of the kinetic energy, spring potential energy and gravitational potential energy initially.
  $K_{f} + U_{s,f} + U_{g,f} = K_{i} + U_{s,i} + U_{g,i} + W$
- $U_{s,i}$  and  $W$  cancel out.
+The work is equal to zero as no work is being done by the surroundings since the system is the Earth, block and spring.
+ $U_{s,i}$  is also equal to zero as there is no initial spring potential energy as the spring is not compressed.
+We then substitute in the equations for kinetic and potential energies.
  $\dfrac{1}{2}mv^2_{f} + mgy_{f} + \dfrac{1}{2}k_{s}s^2_{f} = mgy_{i} + \dfrac{1}{2}mv^2_{i}$
+Multiply across by 2 and divide across by m in order to simplify the equation. Also group like terms.
  $v_{f}^2 = v_{i}^2 + 2g(y_{i} - y_{f}) - \dfrac{k_{s}}{m}{s^2_{f}}$
+Isolate  $v_{f}$  to solve for it:
  $v_{f} = \sqrt {(2 m/s)^2 + 2(9.8 N/kg)(0.5 m) - \dfrac{2000 N/m}{3 kg}{(0.1 m)}^2}$
@@ Line 71: / Line 96: @@
 b:
+We want to find the value for  $y_{f}$  which is the maximum height reached by the bottom of the block above the floor.
 From the Energy Principle:
  $E_{f} = E_{i} + W$
+The total final energy is the sum of the kinetic energy, spring potential energy and gravitational potential energy which is equal to the total initial energy which is the sum of the kinetic energy, spring potential energy and gravitational potential energy initially.
  $K_{f} + U_{s,f} + U_{g,f} = K_{i} + U_{s,i} + U_{g,i} + W$
-$K_{f}, U_{s,f}, U_{s,i}$ and $W$ cancel out.
+The work is equal to zero as no work is being done by the surroundings since the system is the Earth, block and spring.
+$U_{s,i} $and$ U_{s,f}$ are both equal to zero as there is no initial spring potential energy or final spring potential energy as the spring is not compressed in either case.
+$K_{f}$ is also equal to zero as the maximum height is reached when the block has a final kinetic energy of zero.
+Substitute in equations for kinetic and potential energy for remaining terms.
  $mgy_{f} = mgy_{i} + \dfrac{1}{2}mv^2_{i}$
-$y_{f} = y_{i} + \dfrac {v_{i}^2}{2g} = (0.8 m) + \dfrac {2 m/s}^2{2(9.8 N/kg)} = 1.0 m$
+Divide each term by mg.
+$y_{f} = y_{i} + \dfrac {v_{i}^2}{2g}$
+Solve for $y_{f}$
+ $(0.8 m) + \dfrac {(2 m/s)^2}{2(9.8 N/kg)} = 1.0 m$