A horse pulls a 300 kg sled for 50 m over level ground at constant speed, exerting a force of 3000 N at 30° above horizontal. How much work is done by the horse on the sled? 150 kJ 2940 J 130 kJ

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Chapter1: Units, Trigonometry. And Vectors
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**Problem Statement:**

A horse pulls a 300 kg sled for 50 m over level ground at constant speed, exerting a force of 3000 N at 30° above the horizontal. How much work is done by the horse on the sled?

**Options:**

- 150 kJ
- 2940 J
- 130 kJ
- 0 J

**Explanation:**

To find the work done by the horse, we can use the formula:

\[ \text{Work} = F \cdot d \cdot \cos(\theta) \]

Where:
- \( F = 3000 \, \text{N} \) (the force exerted by the horse)
- \( d = 50 \, \text{m} \) (the distance over which the force is applied)
- \( \theta = 30^\circ \) (the angle above the horizontal)

Calculating the component of the force in the direction of the movement:

\[ \text{Work} = 3000 \, \text{N} \times 50 \, \text{m} \times \cos(30^\circ) \]

\[ \text{Work} = 3000 \, \text{N} \times 50 \, \text{m} \times \frac{\sqrt{3}}{2} \]

\[ \text{Work} = 3000 \times 50 \times 0.866 \] 

\[ \text{Work} = 129900 \, \text{J} \]

Since 1 kJ = 1000 J:

\[ \text{Work} = 129.9 \, \text{kJ} \]

Therefore, the closest option is 130 kJ.
Transcribed Image Text:**Problem Statement:** A horse pulls a 300 kg sled for 50 m over level ground at constant speed, exerting a force of 3000 N at 30° above the horizontal. How much work is done by the horse on the sled? **Options:** - 150 kJ - 2940 J - 130 kJ - 0 J **Explanation:** To find the work done by the horse, we can use the formula: \[ \text{Work} = F \cdot d \cdot \cos(\theta) \] Where: - \( F = 3000 \, \text{N} \) (the force exerted by the horse) - \( d = 50 \, \text{m} \) (the distance over which the force is applied) - \( \theta = 30^\circ \) (the angle above the horizontal) Calculating the component of the force in the direction of the movement: \[ \text{Work} = 3000 \, \text{N} \times 50 \, \text{m} \times \cos(30^\circ) \] \[ \text{Work} = 3000 \, \text{N} \times 50 \, \text{m} \times \frac{\sqrt{3}}{2} \] \[ \text{Work} = 3000 \times 50 \times 0.866 \] \[ \text{Work} = 129900 \, \text{J} \] Since 1 kJ = 1000 J: \[ \text{Work} = 129.9 \, \text{kJ} \] Therefore, the closest option is 130 kJ.
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