What is Wg, the work done on the block by the force of gravity as the block moves a distance L up the incline? Express the work done by gravity in terms of the weight w and any other quantities given in the problem introduction.
The Correct Answer and Explanation is:
To calculate the work done on a block by the force of gravity as it moves a distance ( L ) up an incline, we can start by identifying the forces acting on the block and how the work done is defined. The gravitational force acting on the block is represented by its weight, ( w ), which can be expressed as:
[
w = mg
]
where ( m ) is the mass of the block and ( g ) is the acceleration due to gravity. When the block moves up the incline, the gravitational force can be decomposed into two components: one acting perpendicular to the incline and the other acting parallel to it. The component of the gravitational force acting parallel to the incline can be calculated using the angle ( \theta ) of the incline:
[
F_{\text{gravity, parallel}} = w \sin(\theta)
]
The work done by gravity as the block moves up the incline is defined as:
[
W_g = F \cdot d \cdot \cos(\phi)
]
where:
- ( F ) is the force acting on the block (in this case, ( w \sin(\theta) )),
- ( d ) is the distance moved along the incline (( L )),
- ( \phi ) is the angle between the force and the direction of motion. Here, since the gravitational force acts downward while the block moves upward, ( \phi = 180^\circ ) (or (\cos(180^\circ) = -1)).
Substituting in the values, we have:
[
W_g = (w \sin(\theta)) \cdot L \cdot (-1)
]
[
W_g = -w \sin(\theta) \cdot L
]
The negative sign indicates that the work done by gravity is in the opposite direction of the displacement. Therefore, the work done on the block by the force of gravity as it moves a distance ( L ) up the incline is given by:
[
W_g = -wL \sin(\theta)
]
This expression shows that the work done by gravity depends on the weight of the block, the distance moved up the incline, and the angle of the incline.