Let's go through each of the questions step-by-step and solve them.
### Problem 2: Olympic Diver
#### Given:
- Platform height [tex]\( h = 10 \, \text{m} \)[/tex]
- Mass of diver [tex]\( m = 60 \, \text{kg} \)[/tex]
- Gravitational acceleration [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
#### a. Gravitational Potential Energy
Gravitational potential energy (GPE) is given by:
[tex]\[ \text{GPE} = mgh \][/tex]
Substitute the values:
[tex]\[ \text{GPE} = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 10 \, \text{m} \][/tex]
[tex]\[ \text{GPE} = 5880 \, \text{J} \][/tex]
So, the diver has 5880 joules of gravitational potential energy.
#### b. Kinetic Energy as the Diver Strikes the Water
According to the conservation of energy, the gravitational potential energy at the top is converted into kinetic energy (KE) as the diver strikes the water:
[tex]\[ \text{KE} = \text{GPE} \][/tex]
[tex]\[ \text{KE} = 5880 \, \text{J} \][/tex]
So, the kinetic energy as the diver strikes the water is 5880 joules.
#### c. Work Done by the Water on the Diver
The work done by the water to stop the diver is equal to the kinetic energy the diver has as they hit the water:
[tex]\[ \text{Work} = \text{KE} \][/tex]
[tex]\[ \text{Work} = 5880 \, \text{J} \][/tex]
So, the work done by the water is 5880 joules.
#### d. Net Force Applied by the Water
Work done by a force is given by:
[tex]\[ \text{Work} = \text{Force} \times \text{distance} \][/tex]
We can solve for the force:
[tex]\[ \text{Force} = \frac{\text{Work}}{\text{distance}} \][/tex]
Substitute the values (distance the diver stops in water = 2.5 m):
[tex]\[ \text{Force} = \frac{5880 \, \text{J}}{2.5 \, \text{m}} \][/tex]
[tex]\[ \text{Force} = 2352 \, \text{N} \][/tex]
So, the net force applied to the diver by the water is 2352 newtons.
### Problem 3: Spring and Wood Block
#### Given:
- Mass of wood block [tex]\( m = 0.75 \, \text{kg} \)[/tex]
- Compression distance [tex]\( x = 0.2 \, \text{m} \)[/tex]
- Spring constant [tex]\( k = 50 \, \text{N/m} \)[/tex]
#### a. Spring Potential Energy
The potential energy stored in a compressed or stretched spring is given by:
[tex]\[ \text{SPE} = \frac{1}{2} k x^2 \][/tex]
Substitute the values:
[tex]\[ \text{SPE} = \frac{1}{2} \times 50 \, \text{N/m} \times (0.2 \, \text{m})^2 \][/tex]
[tex]\[ \text{SPE} = \frac{1}{2} \times 50 \times 0.04 \][/tex]
[tex]\[ \text{SPE} = \frac{50 \times 0.04}{2} \][/tex]
[tex]\[ \text{SPE} = 1 \, \text{J} \][/tex]
So, the spring potential energy in the system before the wood block begins to move is 1 joule.
### Summary:
2. a. [tex]\( 5880 \, \text{J} \)[/tex] GPE
b. [tex]\( 5880 \, \text{J} \)[/tex] KE
c. [tex]\( 5880 \, \text{J} \)[/tex] Work
d. [tex]\( 2352 \, \text{N} \)[/tex] Force
3. a. [tex]\( 1 \, \text{J} \)[/tex] Spring Potential Energy
