- 1. A 2000 kg loaded coal cart is pulled up at an angle $$ 10^{\circ} $$ inclined at a constant speed by the cable parallel to the incline. The frictional force impeding the motion is 800 N . How much work is done on the cart by the cable as the cart is pulled 30 m . Use $$ g=9.81 \mathrm{~ms} $$.

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To calculate the work done by the cable on the coal cart, we'll follow these steps:

1. **Determine the Tension in the Cable ($$ T $$)**

Since the cart is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance both the frictional force and the component of gravitational force acting down the incline.

$$
   T = F_{\text{friction}} + m \cdot g \cdot \sin(\theta)
   $$

Where:
   - $$ F_{\text{friction}} = 800 \, \text{N} $$
   - $$ m = 2000 \, \text{kg} $$
   - $$ g = 9.81 \, \text{m/s}^2 $$
   - $$ \theta = 10^\circ $$

Calculate the gravitational component:
   
   $$
   \sin(10^\circ) \approx 0.1736
   $$
   
   $$
   m \cdot g \cdot \sin(\theta) = 2000 \times 9.81 \times 0.1736 \approx 3407 \, \text{N}
   $$
   
   Now, calculate $$ T $$:
   
   $$
   T = 800 \, \text{N} + 3407 \, \text{N} = 4207 \, \text{N}
   $$

2. **Calculate the Work Done by the Cable ($$ W $$)**

Work is given by the product of the force and the displacement in the direction of the force.

$$
   W = T \times d
   $$
   
   Where:
   - $$ d = 30 \, \text{m} $$
   
   $$
   W = 4207 \, \text{N} \times 30 \, \text{m} = 126,210 \, \text{J}
   $$

So, the work done by the cable is **126,210 Joules**.

**Final Answer:**  
The cable does 126 210 J of work on the cart as it is pulled 30 meters.
The cable does 126,210 Joules of work on the cart as it is pulled 30 meters.

A 2000 kg loaded coal cart is pulled up at an angle inclined at a
constant speed by the cable parallel to the incline. The frictional force
impeding the motion is 800 N . How much work is done on the cart by
the cable as the cart is pulled 30 m . Use .

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A 2000 kg loaded coal cart is pulled up at an angle inclined at a constant speed by the cable parallel to the incline. The frictional force impeding the motion is 800 N . How much work is done on the cart by the cable as the cart is pulled 30 m . Use .