4. An ant sits on the back of a mouse. The mouse carries the ant across the floor for a distance of 10 meters. Answer the question about the mouse. 5. JD is setting a bucket of water on the ground. 6. You are lifting a 45-newton bag of mulch. Now you are carrying the bag to the garden. You set the bag down in the garden.

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4. An ant sits on the back of a mouse. The mouse carries the ant across the floor for a distance of 10 meters. Answer the question about the mouse. 5. JD is setting a bucket of water on the ground. 6. You are lifting a 45-newton bag of mulch. Now you are carrying the bag to the garden. You set the bag down in the garden.

Answer 1

### Work done by the mouse is dependent on the applied force; Work done in setting the bucket is negligible if gently placed; Lifting the bag requires 45 J of work; Carrying bag horizontally does no work.

Answer 2

## Step 1: Analyze the situation Observe that all sub-questions involve different physical actions and require an understanding of basic physical principles (distance, force, and work). Mainly, we need to determine the relationship between force and work done. ## Step 2: Determine the work done by the mouse Work done, $W$, is defined as the product of force, $F$, and distance, $d$. Assume the force is constant and thus $W = F \cdot d$. Here, $F$ is the force by the mouse against friction, if any, but we can't determine this without further data. Only the distance is given (10 meters). ### Work done by the mouse: $W = F \cdot 10 \text{ meters} $ ## Step 3: Identify the force by placing the bucket When JD sets a bucket of water on the ground, the bucket moves downward until it touches the ground. If JD sets it gently, kinetic energy is converted into other forms of energy due to a negligible distance moved at the final instance, implying no work done horizontally. ### Force involved in setting a bucket of water: $F_{\text{vertical}} = \text{weight of the bucket} $ ## Step 4: Calculate work done lifting the bag When lifting a bag of 45 newtons vertically, the work done, $W$, is given by: $W = F \times d $ where $F$ is 45 N and assuming $d$ is the vertical distance lifted. Let's say up to waist height, maybe 1 meter: ### Work done lifting the bag: $W = 45 \text{ N} \times \text{1 meter} = 45 \text{ J} $ ## Step 5: Analyze carrying the bag Carrying the bag horizontally to the garden implies a horizontal movement where the vertical force (weight) has no displacement component horizontally. Therefore, work done in moving horizontally with constant speed is effectively zero as force perpendicular to direction of displacement yields no work: $W_{\text{horizontal carry}} = 0 \text{ J} $

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