Question

question 5
5 points save answer
a concrete block is pulled 7.0 m across a frictionless surface by means of a rope. the tension in the rope is 40 n; and the net work done on the block is 247 j. what angle does the rope make with the horizontal? the image is not drawn to scale.
88°
28°
47°
62°
41°

Explanation:

Step1: Recall the work formula

The work done \(W\) by a force \(F\) over a displacement \(d\) at an angle \(\theta\) is \(W = Fd\cos\theta\).

Step2: Rearrange the formula to solve for \(\theta\)

We can rewrite the formula as \(\cos\theta=\frac{W}{Fd}\).
Substituting \(W = 247\space J\), \(F = 40\space N\), and \(d = 7.0\space m\) into the formula:
\(\cos\theta=\frac{247}{40\times7.0}=\frac{247}{280}\approx0.882\).

Step3: Find the angle \(\theta\)

Then \(\theta=\cos^{- 1}(0.882)\approx28^{\circ}\).

Answer:

\(28^{\circ}\)