To solve the problem step by step, we will follow these steps:
### Step 1: Identify the given data
- Mass of the man (M) = 70 kg
- Mass of the thrown body (m) = 3 kg
- Velocity of the thrown body (v) = 8 m/s
- Coefficient of friction (μ) = 0.02
### Step 2: Calculate the normal force (N)
The normal force (N) acting on the man is equal to his weight, which can be calculated as:
\[ N = M \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
\[ N = 70 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 686 \, \text{N} \]
### Step 3: Calculate the frictional force (F_f)
The frictional force can be calculated using the formula:
\[ F_f = \mu \cdot N \]
\[ F_f = 0.02 \cdot 686 \, \text{N} = 13.72 \, \text{N} \]
### Step 4: Apply the conservation of momentum
Initially, both the man and the body are at rest, so the initial momentum is zero. After the man throws the body, the final momentum can be expressed as:
\[ 0 = M \cdot V_m + m \cdot v \]
where \( V_m \) is the velocity of the man after throwing the body.
Rearranging gives:
\[ V_m = -\frac{m \cdot v}{M} \]
Substituting the values:
\[ V_m = -\frac{3 \, \text{kg} \cdot 8 \, \text{m/s}}{70 \, \text{kg}} = -\frac{24}{70} \approx -0.343 \, \text{m/s} \]
### Step 5: Calculate the acceleration (a) of the man
The acceleration of the man due to the frictional force can be calculated using Newton's second law:
\[ a = \frac{F_f}{M} \]
\[ a = \frac{13.72 \, \text{N}}{70 \, \text{kg}} \approx 0.196 \, \text{m/s}^2 \]
### Step 6: Use the equations of motion to find the distance (s) slipped
Using the equation of motion:
\[ V^2 = U^2 + 2a s \]
where:
- Final velocity \( V = 0 \) (the man comes to rest)
- Initial velocity \( U = 0.343 \, \text{m/s} \)
- Acceleration \( a = -0.196 \, \text{m/s}^2 \) (deceleration)
Rearranging gives:
\[ 0 = (0.343)^2 + 2(-0.196)s \]
\[ 0.117649 = 0.392s \]
\[ s = \frac{0.117649}{0.392} \approx 0.3 \, \text{m} \]
### Final Answer
The distance the man slips is approximately **0.3 meters**.
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