Question

8 multiple choice 1 point three different forces, a, b, and c act on a box as it slides to the right. each of the forces has the same magnitude, but they each have a different direction. rank the amount of work done by the forces on the box from greatest to least. a = b = c b > c > a c > b = a c > a > b b > a > c

Explanation:

Step1: Recall Work Formula

Work \( W \) is given by \( W = Fd\cos\theta \), where \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between \( \vec{F} \) and \( \vec{d} \). Here, \( F \) (magnitude) and \( d \) are the same for all forces. So \( W \propto \cos\theta \).

Step2: Determine Angles for Each Force

- **Force B**: Direction is same as displacement (\( \theta = 0^\circ \)), so \( \cos(0^\circ) = 1 \).
- **Force A**: Direction is at an acute angle to displacement (so \( 0^\circ < \theta < 90^\circ \)), \( \cos\theta \) is between \( 0 \) and \( 1 \), but less than \( 1 \) (since \( \theta > 0^\circ \)). Wait, no—wait, in diagram A, the force is at an angle below horizontal? Wait, no, the diagram: A has force at an angle (maybe above? Wait, no, the arrow for A is going to the right but angled? Wait, no, looking at the diagrams:
- **Force B**: Horizontal (same direction as displacement, \( \theta = 0^\circ \)).
- **Force A**: At an angle (let's say \( \theta_A \) where \( 0^\circ < \theta_A < 90^\circ \), but wait, no—wait, the displacement is to the right. For force C, the force is vertical downward, so \( \theta = 90^\circ \), so \( \cos(90^\circ) = 0 \). Wait, no:
- **Force C**: Direction is vertical (downward), displacement is horizontal (right), so angle between \( \vec{F} \) and \( \vec{d} \) is \( 90^\circ \), so \( \cos(90^\circ) = 0 \).
- **Force A**: Let's say the force is at an angle \( \theta \) where \( \theta \) is between \( 0^\circ \) and \( 90^\circ \) (but wait, the arrow for A: is it above or below? Wait, the diagram: A has the force coming from the left, angled (maybe downward? No, the arrow is towards the box, angled. Wait, no—displacement is to the right. So for work, the angle between force and displacement:
- **Force B**: \( \theta = 0^\circ \), \( \cos\theta = 1 \).
- **Force A**: The force is at an angle (let's say \( \theta_A \)) where \( \theta_A \) is between \( 0^\circ \) and \( 90^\circ \), but wait, no—if the force is angled towards the right (like a push at an angle), then the horizontal component is \( F\cos\theta \), and vertical component is \( F\sin\theta \). But work depends on the angle between force and displacement. So displacement is right, force A: angle between force and displacement is \( \theta_A \) (acute), so \( \cos\theta_A \) is less than 1 (since \( \theta_A > 0^\circ \))? Wait, no—wait, if the force is horizontal (B), then \( \theta = 0^\circ \), \( \cos\theta = 1 \). If the force is at an angle (A) where the angle between force and displacement is \( \theta_A \) (so \( \theta_A \) is the angle between \( \vec{F} \) and \( \vec{d} \)), then for A, \( \theta_A \) is between \( 0^\circ \) and \( 90^\circ \), so \( \cos\theta_A \) is between \( 0 \) and \( 1 \), but less than 1 (since \( \theta_A > 0^\circ \))? Wait, no—wait, maybe I got A wrong. Wait, no: in diagram A, the force is at an angle (maybe below the horizontal, but still, the angle between force and displacement (right) is \( \theta_A \), so \( \cos\theta_A \) is less than 1 (because \( \theta_A > 0^\circ \)). Wait, no—wait, force B is horizontal (same direction as displacement, \( \theta = 0^\circ \), \( \cos\theta = 1 \)). Force A: the force is at an angle (so \( \theta_A \) is the angle between force and displacement), so \( \cos\theta_A \) is less than 1 (since \( \theta_A > 0^\circ \)). Force C: vertical (downward), displacement is horizontal, so angle between them is \( 90^\circ \), \( \cos(90^\circ) = 0 \). Wait, but that can't be. Wait, no—maybe I mixed up A and C. Wait, no: the problem is to rank work done by the forces. Work is \( Fd\cos\theta \), where \( \theta \) is the angle between force and displacement. So:

- **Force B**: \( \theta = 0^\circ \), \( \cos\theta = 1 \), so \( W_B = Fd(1) \).
- **Force A**: The force is at an angle (let's say \( \theta_A \)) where \( \theta_A \) is between \( 0^\circ \) and \( 90^\circ \), so \( \cos\theta_A \) is between \( 0 \) and \( 1 \), but wait—wait, maybe the force in A is at an angle below the horizontal, but the horizontal component is still contributing. Wait, no—actually, in diagram A, the force is at an angle (maybe above? No, the arrow is going to the box, angled. Wait, maybe the angle for A is such that \( \theta_A \) is less than \( 90^\circ \), so \( \cos\theta_A \) is positive, but less than 1 (since \( \theta_A > 0^\circ \)). Wait, no—force B is horizontal ( \( \theta = 0^\circ \), \( \cos\theta = 1 \) ). Force A: if the force is at an angle (say, \( \theta_A = 30^\circ \) ), then \( \cos(30^\circ) = \sqrt{3}/2 \approx 0.866 \), which is less than 1. Force C: vertical ( \( \theta = 90^\circ \), \( \cos\theta = 0 \) ). Wait, but that would mean \( W_B > W_A > W_C \)? Wait, but the options include B > A > C. Wait, let's recheck:

Wait, the formula is \( W = Fd\cos\theta \), where \( \theta \) is the angle between the force vector and the displacement vector. So:

- For Force B: direction is same as displacement (right), so \( \theta = 0^\circ \), \( \cos(0^\circ) = 1 \). So \( W_B = Fd(1) \).
- For Force A: the force is at an angle (let's say \( \theta_A \)) where \( 0^\circ < \theta_A < 90^\circ \) (since it's angled but still has a horizontal component), so \( \cos\theta_A \) is between \( 0 \) and \( 1 \), so \( W_A = Fd\cos\theta_A \), which is less than \( W_B \) (since \( \cos\theta_A < 1 \)).
- For Force C: direction is vertical (downward), displacement is horizontal (right), so \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \), so \( W_C = Fd(0) = 0 \).

Therefore, the work done is \( W_B > W_A > W_C \), so the ranking from greatest to least is \( B > A > C \).

Answer:

B > A > C (the option corresponding to this ranking)