Question

question 1 (1 point)
a person lifts a pail of water of mass of 1.50 kg from the ground to a deck. 1.00 m above the ground. how much work was done by gravitational force on the pail of water?
a. −14.7 j\t\td.+0.153 j
b. +1.50 j\t\te.+14.7 j
c. −1.50 j

question 2 (1 point)
a ball drops some distance and gains 30 j of kinetic energy. do not ignore air resistance. how much gravitational potential energy did the ball lose?

Explanation:

Question 1

Step1: Recall Work Formula

Work \( W = F \cdot d \cdot \cos\theta \), where \( F \) is force, \( d \) is displacement, \( \theta \) is angle between them. Gravitational force \( F = mg \), \( m = 1.50 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), so \( F = 1.50 \times 9.8 = 14.7 \, \text{N} \) (downward). Displacement \( d = 1.00 \, \text{m} \) (upward). Angle \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \).

Step2: Calculate Work

\( W = 14.7 \times 1.00 \times (-1) = -14.7 \, \text{J} \).

When a ball drops with air resistance, the gravitational potential energy lost (\( \Delta U \)) is converted into kinetic energy (\( \Delta K = 30 \, \text{J} \)) and work done against air resistance (\( W_{\text{air}} \), positive as energy is dissipated). By energy conservation: \( \Delta U = \Delta K + W_{\text{air}} \). Since \( W_{\text{air}} > 0 \), \( \Delta U > 30 \, \text{J} \).

Answer:

a. −14.7 J

Question 2