Select the correct answer.
What is the value of work done on an object when a $0.1 \times 10^2$-newton force moves it 30 meters and the angle between the force and the displacement is $25^{\circ}$?
A. $2.7 \times 10^2$ joules
B. $3.0 \times 10^2$ joules
C. $7.5 \times 10^2$ joules
D. 0 joules
Answer (1)
The problem requires calculating work done given force, displacement, and the angle between them.
The formula for work done is W = F × d × cos ( θ ) .
Substituting the given values, W = 10 × 30 × cos ( 2 5 ∘ ) ≈ 271.89 joules.
Therefore, the work done is approximately 2.7 × 1 0 2 joules .
Explanation
Problem Analysis 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 find the work done on the object.
Work Done Formula 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.
Given Values We are given:
F = 0.1 × 1 0 2 = 10 N
d = 30 m
θ = 2 5 ∘
Calculation Substitute the given values into the formula: W = 10 × 30 × cos ( 2 5 ∘ ) W = 300 × cos ( 2 5 ∘ ) Using a calculator, we find that cos ( 2 5 ∘ ) ≈ 0.9063 . Therefore, W ≈ 300 × 0.9063 = 271.89 joules Rounding to one significant figure, we get W ≈ 2.7 × 1 0 2 joules.
Final Answer Comparing our result with the given options, we see that option A, 2.7 × 1 0 2 joules, is the closest to our calculated value.
Examples
Understanding work done is crucial in many real-world scenarios. For instance, when pushing a lawnmower, only the force component in the direction of motion contributes to the work done. If you push at an angle, some of your force is wasted. Similarly, when pulling a sled, the angle at which you pull affects how much work you do to move the sled forward. Engineers use these principles to design efficient machines and systems, ensuring that energy is used effectively.
