Question

a fourth grader exerts 3.5 newtons of force to push a merry - go - round 11 meters in a circle. calculate the work done by the fourth grader.

Explanation:

Step1: Recall Work Formula

Work \( W \) is done when a force \( F \) acts on an object and displaces it in the direction of the force. The formula is \( W = F \times d \times \cos(\theta) \), where \( \theta \) is the angle between force and displacement. For circular motion (merry - go - round), the force is tangent to the circle, and displacement is also tangent (instantaneous direction). But when moving in a full circle, the total displacement is \( 0 \) (since the object returns to the starting point). Wait, but here the problem says "push a merry - go - round 11 meters in a circle". Wait, maybe it's a misunderstanding. Wait, no—if we consider the work done in circular motion where the force is applied tangentially, but the key point: in circular motion, when an object moves in a circle and returns to the original position, the net displacement is zero. But wait, maybe the problem is considering the distance along the circle as the displacement? No, displacement is a vector. Wait, no—maybe the problem has a mistake, but in the case of work done on a merry - go - round (rotational work), but the formula for translational work is \( W = Fd\cos\theta \). If the force is applied tangentially and we consider the distance along the circle as the path, but displacement is zero. Wait, but maybe the problem is simplified. Wait, no—wait, when you push a merry - go - round, the force is applied tangentially, and the displacement of the point of application (in the direction of the force) is along the tangent. But if it's moving in a circle, after a full circle, the displacement is zero. But the problem says "11 meters in a circle"—maybe it's the arc length, but displacement is zero. Wait, but maybe the problem is not considering the circular motion's displacement (net displacement) but the distance? No, work depends on displacement in the direction of the force. Wait, maybe the problem is a trick question. Let's think again. The formula for work is \( W = F \times d \times \cos(\theta) \), where \( d \) is displacement. If the force is applied in the direction of motion (tangential), and the object moves in a circle, but the net displacement is zero (since it comes back to the start). So \( d = 0 \), so \( W = 3.5\ N\times0\ m\times\cos(\theta)=0\ J \).

Step2: Calculate Work

Given \( F = 3.5\ N \), displacement \( d = 0\ m \) (because it moves in a circle and returns to the original position), so \( W=F\times d\times\cos(\theta) \). Since \( d = 0 \), \( W = 3.5\times0\times\cos(\theta)=0\ J \).

Answer:

\( 0 \) (in Joules)