A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure 6). The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d⃗ of the box is 1.8 m down the inclined plane. What is the work Ww done on the box by the weight of the box?
The Correct Answer and Explanation is:
To find the work ( W_w ) done on the box by the weight of the box as it moves down the inclined plane, let’s break down the problem step by step.
Step 1: Understand the Problem
The box moves down an inclined plane with an angle ( \phi = 30^\circ ) from the horizontal, and its weight is ( w = 2.0 \, \text{N} ). The work done by the weight is essentially the product of the weight component along the direction of displacement and the displacement itself.
Step 2: Resolve the Weight Component along the Inclined Plane
The force of gravity acts vertically downward, so it has a component along the inclined plane and a component perpendicular to it. Only the component along the plane contributes to the work done in this direction.
- The weight component parallel to the plane:
[
w_{\parallel} = w \cdot \sin(\phi)
]
Given:
[
w = 2.0 \, \text{N}, \quad \phi = 30^\circ
]
So:
[
w_{\parallel} = 2.0 \, \text{N} \cdot \sin(30^\circ) = 2.0 \, \text{N} \cdot 0.5 = 1.0 \, \text{N}
]
Step 3: Calculate the Work Done by the Weight
The work done by a force in the direction of displacement is given by:
[
W_w = F_{\parallel} \cdot d
]
where ( F_{\parallel} ) is the component of the weight along the inclined plane, and ( d = 1.8 \, \text{m} ) is the displacement of the box down the plane.
Substituting the values we have:
[
W_w = 1.0 \, \text{N} \cdot 1.8 \, \text{m} = 1.8 \, \text{J}
]
Final Answer
The work ( W_w ) done on the box by its weight as it moves down the inclined plane is 1.8 J.
Explanation
This work calculation focuses on the weight’s contribution along the inclined plane. Since only the component of the weight parallel to the plane affects the box’s movement in that direction, we isolate this part using trigonometry. The angle of inclination (30°) allows us to use ( \sin(30^\circ) = 0.5 ), simplifying the calculation of ( w_{\parallel} ). This component is then multiplied by the displacement, yielding the work done by the gravitational force as it acts along the inclined path.