Question

a fourth grader exerts 7.3 newtons of force to push a merry - go - round 8 meters in a circle. calculate the work done by the fourth grader.
answer
check
what are the units for your answer?
select one:
watts
meters
newtons

Explanation:

Step1: Recall Work Formula

Work \( W \) is force \( F \) times displacement \( d \) in the direction of force: \( W = F \times d \). But when force is perpendicular to displacement (like circular motion, force is radial, displacement tangential), work is zero. Wait, but maybe the problem considers the path length? Wait, no—work is \( F \cdot d \) (dot product), so if force and displacement are perpendicular, work is zero. But let's check: pushing a merry - go - round in a circle—force is tangent? Wait, maybe the problem is simplified. Wait, the force is 7.3 N, distance 8 m. Wait, maybe the problem assumes force and displacement are in the same direction (maybe a mistake, but let's proceed with \( W = F \times d \) if we ignore the circular direction). Wait, no—actually, when moving in a circle, the displacement (change in position) is zero? Wait, no, the problem says "push a merry - go - round 8 meters in a circle"—maybe the distance along the path (arc length) is 8 m. But work done by a force is \( W=\int \vec{F}\cdot d\vec{r} \). If the force is always tangent to the circle (pushing to make it rotate), and the displacement \( d\vec{r} \) is also tangent, then \( \vec{F} \) and \( d\vec{r} \) are in the same direction, so \( W = F\times s \), where \( s \) is the path length. Wait, but maybe the problem has a mistake, but let's check the units. Wait, no—if it's a circle, the displacement (net displacement) is zero, so work should be zero. But maybe the problem is using "distance" as the path length and force in the direction of motion. Wait, let's re - evaluate.

Wait, the key here: in circular motion, if the force is applied tangentially (to make it move along the circle), then the angle between force and displacement (infinitesimal) is zero, so work is \( F\times \) path length. But maybe the problem is a trick question? Wait, no—maybe the problem is wrong, but let's see. Wait, the force is 7.3 N, distance 8 m. If we use \( W = F\times d \) (assuming force and displacement are in the same direction, even though it's a circle—maybe the problem is simplified). Wait, but actually, when moving in a circle, the displacement (final - initial position) is zero, so work should be zero. But maybe the problem means the distance along the path (arc length) and force is tangent, so \( W = 7.3\ N\times8\ m=58.4\ J \)? Wait, but that's incorrect because in circular motion with constant speed, the centripetal force does no work, but a tangential force would do work. Wait, maybe the problem is considering the tangential force. Let's proceed with the formula \( W = F\times d \) (assuming force and displacement are in the same direction, maybe the problem is simplified).

Step2: Calculate Work

Given \( F = 7.3\ N \), \( d = 8\ m \). Using \( W=F\times d \), we get \( W = 7.3\times8=58.4\ J \)? Wait, no—wait, if the motion is circular, the displacement (net) is zero, so work should be zero. But maybe the problem is using "distance" as the path length and force in the direction of motion (tangential). This is a bit confusing. Wait, let's check the basic concept: work is force times displacement in the direction of force. If you push a merry - go - round in a circle, the force is tangent, and the displacement (infinitesimal) is also tangent, so over the path, the work is \( F\times \) path length. But if the problem is considering the net displacement (which is zero, since it's a circle), then work is zero. But the problem says "push a merry - go - round 8 meters in a circle"—maybe "8 meters" is the circumference? No, 8 meters is the distance pushed. Wait, maybe the problem has an error, but let's assume that the force and displacement are in the same direction (linear approximation), so \( W=F\times d \).

Wait, no—actually, when moving in a circle, the displacement vector (from start to end) is zero (since it's a circle, you end up at the start). So \( W=\vec{F}\cdot\vec{d}=Fd\cos\theta \), where \( \theta \) is the angle between force and displacement. If displacement is zero, work is zero. But maybe the problem is using "distance" as the path length (scalar distance) instead of displacement (vector). In that case, if we consider the work done by a tangential force over a path length \( s \), then \( W = F\times s \), because the force and the direction of motion (tangent) are in the same direction, so \( \cos\theta = 1 \) for each infinitesimal displacement. So \( W=7.3\ N\times8\ m = 58.4\ J \). But wait, the units of work are Joules (N·m). But let's check the second part: the units of work are Joules, but the options are Watts (power), meters (length), Newtons (force). Wait, that's a mistake in the options? Wait, no—maybe I misread. Wait, the first question is to calculate work, the second is units. But the options are Watts, meters, Newtons. Wait, that can't be. Wait, maybe the first question is power? No, the question says "work done". Wait, maybe the problem is wrong, but let's proceed.

Wait, no—maybe I made a mistake. Let's re - start.

Step1: Recall Work Definition

Work is done when a force causes displacement in its direction. The formula is \( W = F\times d \), where \( F \) is force, \( d \) is displacement in the direction of force. In circular motion, if the object moves in a circle, the displacement (change in position) is zero (since it returns to the start). So \( W = F\times d\times\cos\theta \), where \( d = 0 \) (displacement), so \( W = 0 \).

Step2: Calculate Work

Given \( F = 7.3\ N \), displacement \( d = 0 \) (since it's a circle, start and end position are the same), so \( W=7.3\ N\times0\ m\times\cos\theta = 0\ J \).

For the units question: work is measured in Joules, but the options are wrong. But maybe the problem intended to ask for something else. But based on the first calculation, if we consider displacement as zero, work is zero.

Answer:

For the work calculation: \( 0 \) (if displacement is zero) or \( 58.4\ J \) (if path length is used). But considering the displacement (vector) is zero, the work done is \( 0 \). For the units, since work is in Joules, but the options are incorrect. However, if we follow the first calculation (path length), work is \( 58.4\ J \), but the units options are wrong. But maybe the problem has a mistake. But based on the physics concept, when moving in a circle (displacement zero), work is zero.

(Note: There seems to be an error in the units options provided, as the unit of work is Joule, not among the given options. But for the work calculation, if we consider the displacement is zero (since it's a circle, starting and ending at the same point), the work done is \( 0 \).)