Select the correct answer.
What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and the displacement is 25°?
A. 27 x 10² joules
B. 3.0 x 10² joules
C. 7.5 x 10² joules
D. 0 joules
Answer (2)
The work done is calculated using the formula W = F ⋅ d ⋅ cos ( θ ) .
Substitute the given values: F = 10 N, d = 30 m, and θ = 2 5 ∘ .
Calculate the work done: W = 10 × 30 × cos ( 2 5 ∘ ) ≈ 271.89 Joules.
Approximate the result to match the given options: W ≈ 2.7 × 1 0 2 Joules, so the answer is 2.7 × 1 0 2 joules .
Explanation
Understanding the Problem We are given a force acting on an object, the distance the object moves, and the angle between the force and the displacement. We need to calculate the work done on the object.
Recalling the Formula for Work Done The formula for work done is given by: W = F ⋅ d ⋅ cos ( θ ) where:
W is the work done,
F is the magnitude of the force,
d is the magnitude of the displacement, and
θ is the angle between the force and the displacement.
Substituting the Given Values We are given:
F = 0.1 × 1 0 2 = 10 N
d = 30 m
θ = 2 5 ∘ Substituting these values into the formula, we get: W = 10 × 30 × cos ( 2 5 ∘ ) W = 300 × cos ( 2 5 ∘ )
Calculating the Work Done We know that cos ( 2 5 ∘ ) ≈ 0.9063 . Therefore, W = 300 × 0.9063 ≈ 271.89 Joules Since we need to select the correct answer from the given options, we can approximate this value to 270 Joules = 2.7 × 1 0 2 Joules .
Selecting the Correct Answer Comparing our calculated value with the given options, we see that option A, 2.7 × 1 0 2 joules, is the closest to our calculated value. Therefore, the correct answer is A.
Examples
Understanding work done is crucial in many real-world scenarios. For instance, when pushing a lawn mower, only the force component in the direction of motion contributes to the work done. Similarly, when pulling a suitcase with wheels, the angle at which you pull affects the amount of work you do. Calculating work helps engineers design efficient machines and understand energy transfer in various systems.
Using the formula for work, we calculated that the work done on the object is approximately 2.7 × 1 0 2 joules. This corresponds to option A in the multiple-choice answers. The calculation involved substituting the values of force, distance, and angle into the work formula and approximating the result.
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