To solve the problem, we will follow these steps:
### Step 1: Understand the Forces Acting on the Crate
The crate has a weight (force due to gravity) of 80 N acting vertically downward. The man is pushing the crate up a frictionless slope at an angle of 30 degrees to the horizontal.
### Step 2: Calculate the Mass of the Crate
We know that weight (W) is given by the formula:
\[ W = m \cdot g \]
Where:
- \( W = 80 \, \text{N} \) (weight of the crate)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity, approximate value)
To find the mass (m):
\[ m = \frac{W}{g} = \frac{80 \, \text{N}}{10 \, \text{m/s}^2} = 8 \, \text{kg} \]
### Step 3: Determine the Acceleration of the Crate
The problem states that the speed of the crate increases at a rate of 1.5 m/s². This means:
\[ a = 1.5 \, \text{m/s}^2 \]
### Step 4: Analyze the Forces Along the Slope
The force exerted by the man (F) must overcome both the component of the weight acting down the slope and provide the necessary acceleration. The component of the weight acting down the slope can be calculated as:
\[ W_{\text{parallel}} = W \cdot \sin(\theta) \]
Where:
- \( \theta = 30^\circ \)
Calculating \( W_{\text{parallel}} \):
\[ W_{\text{parallel}} = 80 \, \text{N} \cdot \sin(30^\circ) = 80 \, \text{N} \cdot \frac{1}{2} = 40 \, \text{N} \]
### Step 5: Apply Newton's Second Law
According to Newton's second law:
\[ F - W_{\text{parallel}} = m \cdot a \]
Substituting the known values:
\[ F - 40 \, \text{N} = 8 \, \text{kg} \cdot 1.5 \, \text{m/s}^2 \]
\[ F - 40 \, \text{N} = 12 \, \text{N} \]
\[ F = 12 \, \text{N} + 40 \, \text{N} = 52 \, \text{N} \]
### Step 6: Calculate the Work Done by the Man
The work done (W) by the man is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Since the force is parallel to the displacement, the angle \( \theta = 0 \) degrees, and thus \( \cos(0) = 1 \):
\[ W = F \cdot d \]
Where:
- \( d = 5 \, \text{m} \) (distance moved)
Substituting the values:
\[ W = 52 \, \text{N} \cdot 5 \, \text{m} = 260 \, \text{J} \]
### Final Answer
The work done by the man is **260 Joules**.
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