Question

question 3 (1 point) saved
a container of water is lifted vertically 3.0 m then returned to its original position. if the total force is 30 n, how much work was done?
45 j
90 j
180 j
no work was done.
question 4 (1 point) saved

Explanation:

Step1: Recall Work Formula

Work \( W = F \cdot d \cdot \cos\theta \), where \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between force and displacement.

Step2: Analyze Displacement

The container is lifted 3.0 m up (displacement \( d_1 = 3.0 \, \text{m} \), \( \theta_1 = 0^\circ \), \( \cos 0^\circ = 1 \)) and then lowered 3.0 m down (displacement \( d_2 = 3.0 \, \text{m} \), \( \theta_2 = 180^\circ \), \( \cos 180^\circ = -1 \)).

Step3: Calculate Work for Each Part

• Work up: \( W_1 = F \cdot d_1 \cdot \cos 0^\circ = 30 \, \text{N} \cdot 3.0 \, \text{m} \cdot 1 = 90 \, \text{J} \)
• Work down: \( W_2 = F \cdot d_2 \cdot \cos 180^\circ = 30 \, \text{N} \cdot 3.0 \, \text{m} \cdot (-1) = -90 \, \text{J} \)

Step4: Total Work

Total work \( W_{\text{total}} = W_1 + W_2 = 90 \, \text{J} + (-90 \, \text{J}) = 0 \, \text{J} \) (so no net work done).

Answer:

No work was done.