To solve the problem, we will follow these steps:
### Step 1: Determine the work done by gravity
The work done by gravity (Wg) when the ball falls from a height of 5 m to a depth of 2 m underwater can be calculated using the formula:
\[ W_g = m \cdot g \cdot h \]
where:
- \( m = 1 \, \text{kg} \) (mass of the ball)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 7 \, \text{m} \) (total distance fallen, which is 5 m above water plus 2 m underwater)
Calculating:
\[ W_g = 1 \cdot 10 \cdot 7 = 70 \, \text{J} \]
### Step 2: Determine the buoyant force
The buoyant force (Fb) acting on the ball can be calculated using Archimedes' principle:
\[ F_b = \rho_w \cdot V \cdot g \]
where:
- \( \rho_w \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( V \) is the volume of the ball, which can be expressed as \( \frac{m}{\rho_s} \) where \( \rho_s \) is the density of the ball.
Given the relative density \( s = \frac{2}{3} \), we can express the density of the ball as:
\[ \rho_s = \frac{2}{3} \cdot \rho_w = \frac{2}{3} \cdot 1000 = \frac{2000}{3} \, \text{kg/m}^3 \]
Now, the volume \( V \) of the ball is:
\[ V = \frac{m}{\rho_s} = \frac{1}{\frac{2000}{3}} = \frac{3}{2000} \, \text{m}^3 \]
Now, substituting into the buoyant force equation:
\[ F_b = 1000 \cdot \frac{3}{2000} \cdot 10 = 15 \, \text{N} \]
### Step 3: Calculate the work done by the buoyant force
The work done by the buoyant force (Wb) over the distance of 2 m is:
\[ W_b = F_b \cdot d \]
where \( d = 2 \, \text{m} \):
\[ W_b = 15 \cdot 2 = 30 \, \text{J} \]
Since the buoyant force acts upwards while the ball moves downwards, the work done by buoyant force is negative:
\[ W_b = -30 \, \text{J} \]
### Step 4: Apply the work-energy principle
According to the work-energy theorem:
\[ W_g + W_b + W_r = \Delta KE \]
where \( W_r \) is the work done by the resistive forces and \( \Delta KE \) is the change in kinetic energy. Since the ball starts from rest and comes to rest, the change in kinetic energy is zero:
\[ 70 - 30 + W_r = 0 \]
### Step 5: Solve for the work done by resistive forces
Rearranging the equation:
\[ W_r = -70 + 30 = -40 \, \text{J} \]
### Step 6: Relate work done by resistive forces to the given expression
We are given that the work done by the resistive forces is \( -10n \):
\[ -10n = -40 \]
Solving for \( n \):
\[ n = \frac{40}{10} = 4 \]
### Final Answer
The value of \( n \) is \( 4 \).
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