Answer :
Certainly! Let's solve the problem step-by-step.
### Problem Statement
We're given:
- A 2.40-kg block is dropped from a height of [tex]\( 6.00 \, \text{m} \)[/tex].
- The acceleration due to gravity ([tex]\( g \)[/tex]) is [tex]\( 9.81 \, \text{m/s}^2 \)[/tex].
- When the block compresses the spring by [tex]\( 15.0 \, \text{cm} \)[/tex] (or [tex]\( 0.15 \, \text{m} \)[/tex]), it is momentarily at rest.
- We need to find the speed of the block when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex] (or [tex]\( 0.05 \, \text{m} \)[/tex]).
### Step-by-Step Solution
1. Calculate the initial potential energy of the block:
The potential energy (PE) when the block is at height [tex]\( h \)[/tex] is given by:
[tex]\[ \text{PE}_{\text{initial}} = m \cdot g \cdot h \][/tex]
Substituting the values:
[tex]\[ \text{PE}_{\text{initial}} = 2.40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 6.00 \, \text{m} = 141.264 \, \text{J} \][/tex]
2. Calculate the spring constant [tex]\( k \)[/tex]:
When the block is at rest after compressing the spring by [tex]\( 0.15 \, \text{m} \)[/tex], the potential energy due to gravity is completely stored in the spring. At this point:
[tex]\[ 0.5 \cdot k \cdot (\Delta x_{\text{rest}})^2 = m \cdot g \cdot h \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{2 \cdot m \cdot g \cdot h}{(\Delta x_{\text{rest}})^2} \][/tex]
Substituting the values:
[tex]\[ k = \frac{2 \cdot 2.40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 6.00 \, \text{m}}{(0.15 \, \text{m})^2} = 6278.400 \, \text{N/m} \][/tex]
3. Calculate the potential energy stored in the spring when compressed by [tex]\( 5.0 \, \text{cm} \)[/tex]:
The potential energy in the spring (PE_spring) when compressed by [tex]\( \Delta x_{\text{target}} = 0.05 \, \text{m} \)[/tex]:
[tex]\[ \text{PE}_{\text{spring}, \text{target}} = 0.5 \cdot k \cdot (\Delta x_{\text{target}})^2 \][/tex]
Substituting:
[tex]\[ \text{PE}_{\text{spring}, \text{target}} = 0.5 \cdot 6278.400 \, \text{N/m} \cdot (0.05 \, \text{m})^2 = 7.848 \, \text{J} \][/tex]
4. Calculate the kinetic energy when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex]:
Using the conservation of energy, the total energy initially present (which is the gravitational potential energy) is converted to both kinetic energy (KE) and the spring's potential energy.
Total initial energy:
[tex]\[ \text{E}_{\text{total, initial}} = \text{PE}_{\text{initial}} \][/tex]
The kinetic energy when the spring is compressed by [tex]\( 0.05 \, \text{m} \)[/tex]:
[tex]\[ \text{KE}_{\text{target}} = \text{E}_{\text{total, initial}} - \text{PE}_{\text{spring}, \text{target}} \][/tex]
Substituting the values:
[tex]\[ \text{KE}_{\text{target}} = 141.264 \, \text{J} - 7.848 \, \text{J} = 133.416 \, \text{J} \][/tex]
5. Calculate the speed of the block at [tex]\( 0.05 \, \text{m} \)[/tex] compression:
The kinetic energy is related to the speed [tex]\( v \)[/tex] by:
[tex]\[ \text{KE} = 0.5 \cdot m \cdot v^2 \][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \cdot \text{KE}}{m}} \][/tex]
Substituting the values:
[tex]\[ v = \sqrt{\frac{2 \cdot 133.416 \, \text{J}}{2.40 \, \text{kg}}} = 10.544 \, \text{m/s} \][/tex]
### Final Answer
The speed of the block when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex] is approximately [tex]\( 10.544 \, \text{m/s} \)[/tex].
### Problem Statement
We're given:
- A 2.40-kg block is dropped from a height of [tex]\( 6.00 \, \text{m} \)[/tex].
- The acceleration due to gravity ([tex]\( g \)[/tex]) is [tex]\( 9.81 \, \text{m/s}^2 \)[/tex].
- When the block compresses the spring by [tex]\( 15.0 \, \text{cm} \)[/tex] (or [tex]\( 0.15 \, \text{m} \)[/tex]), it is momentarily at rest.
- We need to find the speed of the block when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex] (or [tex]\( 0.05 \, \text{m} \)[/tex]).
### Step-by-Step Solution
1. Calculate the initial potential energy of the block:
The potential energy (PE) when the block is at height [tex]\( h \)[/tex] is given by:
[tex]\[ \text{PE}_{\text{initial}} = m \cdot g \cdot h \][/tex]
Substituting the values:
[tex]\[ \text{PE}_{\text{initial}} = 2.40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 6.00 \, \text{m} = 141.264 \, \text{J} \][/tex]
2. Calculate the spring constant [tex]\( k \)[/tex]:
When the block is at rest after compressing the spring by [tex]\( 0.15 \, \text{m} \)[/tex], the potential energy due to gravity is completely stored in the spring. At this point:
[tex]\[ 0.5 \cdot k \cdot (\Delta x_{\text{rest}})^2 = m \cdot g \cdot h \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{2 \cdot m \cdot g \cdot h}{(\Delta x_{\text{rest}})^2} \][/tex]
Substituting the values:
[tex]\[ k = \frac{2 \cdot 2.40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 6.00 \, \text{m}}{(0.15 \, \text{m})^2} = 6278.400 \, \text{N/m} \][/tex]
3. Calculate the potential energy stored in the spring when compressed by [tex]\( 5.0 \, \text{cm} \)[/tex]:
The potential energy in the spring (PE_spring) when compressed by [tex]\( \Delta x_{\text{target}} = 0.05 \, \text{m} \)[/tex]:
[tex]\[ \text{PE}_{\text{spring}, \text{target}} = 0.5 \cdot k \cdot (\Delta x_{\text{target}})^2 \][/tex]
Substituting:
[tex]\[ \text{PE}_{\text{spring}, \text{target}} = 0.5 \cdot 6278.400 \, \text{N/m} \cdot (0.05 \, \text{m})^2 = 7.848 \, \text{J} \][/tex]
4. Calculate the kinetic energy when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex]:
Using the conservation of energy, the total energy initially present (which is the gravitational potential energy) is converted to both kinetic energy (KE) and the spring's potential energy.
Total initial energy:
[tex]\[ \text{E}_{\text{total, initial}} = \text{PE}_{\text{initial}} \][/tex]
The kinetic energy when the spring is compressed by [tex]\( 0.05 \, \text{m} \)[/tex]:
[tex]\[ \text{KE}_{\text{target}} = \text{E}_{\text{total, initial}} - \text{PE}_{\text{spring}, \text{target}} \][/tex]
Substituting the values:
[tex]\[ \text{KE}_{\text{target}} = 141.264 \, \text{J} - 7.848 \, \text{J} = 133.416 \, \text{J} \][/tex]
5. Calculate the speed of the block at [tex]\( 0.05 \, \text{m} \)[/tex] compression:
The kinetic energy is related to the speed [tex]\( v \)[/tex] by:
[tex]\[ \text{KE} = 0.5 \cdot m \cdot v^2 \][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \cdot \text{KE}}{m}} \][/tex]
Substituting the values:
[tex]\[ v = \sqrt{\frac{2 \cdot 133.416 \, \text{J}}{2.40 \, \text{kg}}} = 10.544 \, \text{m/s} \][/tex]
### Final Answer
The speed of the block when the spring is compressed by [tex]\( 5.0 \, \text{cm} \)[/tex] is approximately [tex]\( 10.544 \, \text{m/s} \)[/tex].
