A tractor pulls a log with a mass of 500 kg along the ground for 100 m. The rope( between the tractor and the log) makes an angle of 30°with the ground and is acted on by a tensile force of 5000N. How much work does the tractor perform in this scenario? (sin 30°=0.5 , cos 30°=0.866,tan30°=0.577)
Correct answer: C. 433kJ
- A. 250 kJ
- B. 289 kJ
- C. 433kJ
- D. 500 kJ
Explanation
The work done by the tractor is calculated using the formula: W = F * d * cos(θ), where F is the force, d is the distance, and θ is the angle between the force and the direction of motion. In this case, the tensile force is 5000 N, the distance is 100 m, and the angle θ is 30°. Thus, the work done is:W = 5000 N * 100 m * cos(30°) = 5000 * 100 * 0.866 = 433,000 J, or 433 kJ.Option C is correct because it correctly applies the cosine component of the force in the direction of displacement. Option A and B miscalculate the effect of the angle, while Option D assumes the force is completely aligned with the displacement, ignoring the angle.
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