Question

a 17.0-kg crate is to be pulled a distance of 20.0 m, requiring 1210 j of work to be done on the crate. the job is done by attaching a rope and pulling with a force of 75.0 n. at what angle is the rope held?

Explanation:

Step1: Recall the work formula

The formula for work done by a force is \( W = Fd\cos\theta \), where \( W \) is work, \( F \) is force, \( d \) is distance, and \( \theta \) is the angle between the force and displacement.

Step2: Rearrange for \( \cos\theta \)

We need to solve for \( \theta \), so first rearrange the formula: \( \cos\theta=\frac{W}{Fd} \)

Step3: Substitute values

Given \( W = 1210\space J \), \( F = 75.0\space N \), \( d = 20.0\space m \). Substitute into the formula:
\( \cos\theta=\frac{1210}{75.0\times20.0} \)
Calculate the denominator: \( 75.0\times20.0 = 1500 \)
Then \( \cos\theta=\frac{1210}{1500}\approx0.8067 \)

Step4: Find \( \theta \)

Take the arccosine: \( \theta=\arccos(0.8067) \)
Using a calculator, \( \theta\approx36.3^\circ \)

Answer:

The rope is held at an angle of approximately \( \boldsymbol{36.3^\circ} \) (or depending on calculator precision, around \( 36^\circ \) to \( 36.5^\circ \)).