Question

this final exam is cumulative and covers material from the entire course.
instructions from the list of choices, select the one best answer.
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question 31 of 62
1.6 points save answer
a rope is tied to a 0.50 kg box and used to pull a box 2.5 m along a horizontal floor. the rope makes an angle of 42° with the horizontal and has a tension of 10 n. the opposing frictional force between the box and the floor is 1.8 n.
how much work is done by the tension in the rope?
29 j
17 j
19 j
25 j

Explanation:

Step1: Recall Work Formula for Force at Angle

The work done by a force \( F \) over a displacement \( d \) at an angle \( \theta \) to the displacement is given by \( W = Fd\cos\theta \). Here, the tension \( F = 10\,\text{N} \), displacement \( d = 2.5\,\text{m} \), and angle \( \theta = 42^\circ \).

Step2: Calculate Cosine of Angle

First, find \( \cos(42^\circ) \). Using a calculator, \( \cos(42^\circ) \approx 0.7431 \).

Step3: Substitute Values into Formula

Substitute \( F = 10\,\text{N} \), \( d = 2.5\,\text{m} \), and \( \cos(42^\circ) \approx 0.7431 \) into \( W = Fd\cos\theta \):
\( W = 10 \times 2.5 \times 0.7431 \)
\( W = 25 \times 0.7431 \approx 18.5775\,\text{J} \), which is approximately \( 19\,\text{J} \) (rounding to a reasonable value for the options).

Answer:

19 J (corresponding to the option "19 J")