To determine the work done by the force of gravity on the 5.00 kg block as it slides down a 30.0-degree incline, we can use the formula for work done by a force, which is given by:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where:
- \( W \) is the work done,
- \( F \) is the force acting in the direction of movement,
- \( d \) is the distance moved,
- \( \theta \) is the angle between the force and the direction of movement.
Here, the force of gravity acting on the block can be found by:
- Calculate the gravitational force:
\[ F_{\text{gravity}} = m \cdot g \] Where:
- \( m = 5.00 , \text{kg} \) (mass of the block),
- \( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity).
So,
\[ F_{\text{gravity}} = 5.00 , \text{kg} \times 9.81 , \text{m/s}^2 = 49.05 , \text{N} \]
- Calculate the component of gravitational force acting down the incline:
The component of the gravitational force acting parallel to the incline is given by:
\[ F_{\text{parallel}} = F_{\text{gravity}} \cdot \sin(\theta) \]
For a 30-degree incline:
\[ \sin(30^\circ) = 0.5 \]
Thus,
\[ F_{\text{parallel}} = 49.05 , \text{N} \cdot 0.5 = 24.525 , \text{N} \]
- Calculate the work done by gravity:
The distance down the incline is given as \( d = 2.50 , \text{m} \). Since the gravitational force component acts in the same direction as the movement, the angle \( \theta \) between the gravitational force down the incline and the direction of the movement is 0 degrees. Therefore, \( \cos(0) = 1 \).
Now we can calculate the work done:
\[ W = F_{\text{parallel}} \cdot d \cdot \cos(0^\circ) = F_{\text{parallel}} \cdot d \] \[ W = 24.525 , \text{N} \cdot 2.50 , \text{m} = 61.3125 , \text{J} \]
Thus, the work done by the force of gravity on the block as it slides down the incline is approximately:
\[ W \approx 61.31 , \text{J} \]