Understanding Work and Force
In physics, work is defined as the product of the component of force acting in the direction of displacement and the magnitude of that displacement. Mathematically, it is expressed as:
$$W = F \cdot d \cdot \cos(\theta)$$
Where:
- $W$ is the work done.
- $F$ is the magnitude of the applied force.
- $d$ is the displacement.
- $\theta$ is the angle between the force vector and the displacement vector.
Step-by-Step Solution
1. Identify Given Data
- Weight of the block ($W_g$) = $150\text{ N}$
- Coefficient of kinetic friction ($\mu_k$) = $0.20$
- Displacement ($d$) = $20\text{ m}$
- Angle with vertical = $60^\circ$. Therefore, the angle with the horizontal surface ($\theta$) = $90^\circ - 60^\circ = 30^\circ$.
2. Analyze Forces
The block moves at constant velocity, meaning net force in both x and y directions is zero.
- Vertical equilibrium: $N + F\sin(\theta) = W_g$ => $N = W_g - F\sin(30^\circ)$
- Horizontal equilibrium: $F\cos(30^\circ) = f_k$, where $f_k = \mu_k \cdot N$
Substitute $N$ into the friction equation: $F\cos(30^\circ) = \mu_k \cdot (W_g - F\sin(30^\circ))$ $F(0.866) = 0.20 \cdot (150 - F(0.5))$ $0.866F = 30 - 0.1F$ $0.966F = 30$ $F \approx 31.06\text{ N}$
3. Calculate Work Done
$W = F \cdot d \cdot \cos(30^\circ)$ $W = 31.06 \cdot 20 \cdot 0.866$ $W \approx 538.08\text{ J}$