QUESTION IMAGE
Question
physics i
- what is work? what is the unit of work? define this unit in terms of kg, m, and s.
- you apply a force of 100 n to a large box and push it a distance of 5.0 m. how much work have you done on the box?
- you apply a force of 100 n to a large box, which does not move. how much work have you done on the box?
- the box in question 3 weighs 50 n. if it doesnt move up or down, how much work is done on the box by gravity?
- how much work would be required to lift a 50 n box to a height of 5.0 m?
- a crane at the docks is capable of doing 550 j of work. how high can it lift a load that weighs 45 n?
- how much work is done by any centripetal force? explain
Question 2 Solution:
Step1: Recall Work Formula
Work \( W \) is calculated as \( W = F \cdot d \), where \( F \) is force and \( d \) is distance (when force is parallel to displacement).
Given \( F = 100 \, \text{N} \), \( d = 5.0 \, \text{m} \).
Step2: Substitute Values
Substitute into the formula: \( W = 100 \, \text{N} \times 5.0 \, \text{m} \).
Step3: Calculate Result
\( W = 500 \, \text{J} \) (since \( 1 \, \text{N} \cdot \text{m} = 1 \, \text{J} \)).
Question 3 Solution:
Step1: Recall Work Condition
Work requires displacement (\( d = 0 \) here, as the box doesn’t move).
Step2: Apply Work Formula
Using \( W = F \cdot d \), substitute \( d = 0 \, \text{m} \).
Step3: Determine Work
\( W = 100 \, \text{N} \times 0 \, \text{m} = 0 \, \text{J} \).
Question 4 Solution:
Step1: Analyze Gravity and Displacement
Gravity acts vertically, but the box has no vertical displacement (\( d = 0 \) in the vertical direction).
Step2: Apply Work Formula
Using \( W = F \cdot d \), with \( F = 50 \, \text{N} \) (weight) and \( d = 0 \, \text{m} \).
Step3: Calculate Work
\( W = 50 \, \text{N} \times 0 \, \text{m} = 0 \, \text{J} \).
Question 5 Solution:
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- Work: Energy transferred by a force acting through displacement (\( W = F \cdot d \cdot \cos\theta \), \( \theta \) = angle between force and displacement).
- Unit of Work: Joule (J).
- Joule in SI Units: \( 1 \, \text{J} = 1 \, \text{N} \cdot \text{m} = 1 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \) (since \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m}/\text{s}^2 \), so \( 1 \, \text{N} \cdot \text{m} = 1 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \)).
Final Answers (by Question):
- Work: Energy via force-displacement; Unit: Joule (J); \( 1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \).
- \( \boldsymbol{500 \, \text{J}} \)
- \( \boldsymbol{0 \, \text{J}} \)
- \( \boldsymbol{0 \, \text{J}} \)
- \( \boldsymbol{250 \, \text{J}} \)
- \( \boldsymbol{\approx 12.2 \, \text{m}} \) (or \( \frac{110}{9} \approx 12.22 \, \text{m} \))
- \( \boldsymbol{0 \, \text{J}} \) (because centripetal force is perpendicular to displacement, \( \cos 90^\circ = 0 \)).