Question

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)
an object is released from rest a height h above the ground. a second object with four times the mass of the first if released from the same height. the potential energy of the second object compared to the first is
one - fourth as much.
one - half as much.
twice as much.
four times as much.

Explanation:

First Question (Work Done)

Step1: Recall Work Formula

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

Step2: Analyze Displacement

The container is lifted 3.0 m then returned to original position. Total displacement \( d = 3.0 - 3.0 = 0 \, \text{m} \).

Step3: Calculate Work

Substitute \( d = 0 \) into \( W = F \cdot d \cdot \cos\theta \). \( W = 30 \cdot 0 \cdot \cos\theta = 0 \, \text{J} \). So no work is done.

Step1: Recall PE Formula

Gravitational potential energy \( PE = mgh \), where \( m \) is mass, \( g \) is gravity, \( h \) is height.

Step2: Compare PE of Two Objects

Let first object have mass \( m_1 = m \), height \( h_1 = h \). Second object: \( m_2 = 4m \), \( h_2 = h \).
\( PE_1 = mgh \), \( PE_2 = 4mgh \).
\( \frac{PE_2}{PE_1} = \frac{4mgh}{mgh} = 4 \). So \( PE_2 \) is 4 times \( PE_1 \).

Answer:

No work was done.

Second Question (Gravitational Potential Energy)