Worksheet on work, energy, and power featuring physics problems with diagrams.
A worksheet titled "Work, energy and power" with five questions involving physics problems related to work, energy, and power, including diagrams of inclined planes and forces acting on objects.
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Step-by-step solution for: Free Collection of Work, Power, and Energy Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Collection of Work, Power, and Energy Worksheets
1.1
a. The work done by gravity is calculated using the formula:
\( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the component of the force in the direction of displacement.
The gravitational force component parallel to the incline is \( mg \sin(\theta) \).
Here, \( m = 60 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( \theta = 40^\circ \), and \( d = 3 \, \text{m} \).
So, \( W = (60 \cdot 9.8 \cdot \sin(40^\circ)) \cdot 3 \).
\( \sin(40^\circ) \approx 0.6428 \), so:
\( W = (60 \cdot 9.8 \cdot 0.6428) \cdot 3 \approx (378.7) \cdot 3 \approx 1136.1 \, \text{J} \).
The work done by gravity is approximately \( 1136 \, \text{J} \).
b. The normal force is perpendicular to the displacement (which is along the incline). Therefore, the angle between the normal force and displacement is \( 90^\circ \), and \( \cos(90^\circ) = 0 \).
So, the work done by the normal force is \( 0 \, \text{J} \).
1.2
a. The work done by gravity is zero because the displacement is horizontal and gravity acts vertically. The angle between the force of gravity and displacement is \( 90^\circ \), so \( W = F \cdot d \cdot \cos(90^\circ) = 0 \).
The work done by gravity is \( 0 \, \text{J} \).
b. The work done by the applied force is calculated using \( W = F \cdot d \cdot \cos(\theta) \), where \( F = 60 \, \text{N} \), \( d = 3.25 \, \text{m} \), and \( \theta = 30^\circ \).
\( W = 60 \cdot 3.25 \cdot \cos(30^\circ) \).
\( \cos(30^\circ) \approx 0.866 \), so:
\( W = 60 \cdot 3.25 \cdot 0.866 \approx 171.9 \, \text{J} \).
The work done by the applied force is approximately \( 172 \, \text{J} \).
1.3
a. Free body diagram:
- Gravitational force (\( mg \)) acting vertically downward.
- Normal force (\( N \)) acting perpendicular to the incline.
- Applied force (\( F_{\text{applied}} = 8000 \, \text{N} \)) acting parallel to the incline upward.
- Frictional force (\( f = 20 \, \text{N} \)) acting parallel to the incline downward (opposite to motion).
b. The net work done on the car is the sum of the work done by each force.
- Work done by gravity: \( W_{\text{gravity}} = -mg \cdot d \cdot \sin(\theta) \) (negative because it opposes motion up the incline).
\( m = 1200 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( d = 3 \, \text{m} \), \( \theta = 30^\circ \).
\( W_{\text{gravity}} = -1200 \cdot 9.8 \cdot 3 \cdot \sin(30^\circ) = -1200 \cdot 9.8 \cdot 3 \cdot 0.5 = -17640 \, \text{J} \).
- Work done by normal force: \( 0 \, \text{J} \) (perpendicular to displacement).
- Work done by applied force: \( W_{\text{applied}} = 8000 \cdot 3 = 24000 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -20 \cdot 3 = -60 \, \text{J} \).
Net work: \( W_{\text{net}} = -17640 + 0 + 24000 - 60 = 6300 \, \text{J} \).
The net work done on the car is \( 6300 \, \text{J} \).
1.4
Using the work-energy principle: Net work done = change in kinetic energy.
Net work \( W_{\text{net}} = W_{\text{applied}} + W_{\text{friction}} \).
- Work done by applied force: \( W_{\text{applied}} = F \cdot d \cdot \cos(\theta) = 50 \cdot 2.60 \cdot \cos(30^\circ) \).
\( \cos(30^\circ) \approx 0.866 \), so \( W_{\text{applied}} = 50 \cdot 2.60 \cdot 0.866 \approx 112.58 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -f \cdot d = -4.2 \cdot 2.60 = -10.92 \, \text{J} \).
Net work: \( W_{\text{net}} = 112.58 - 10.92 = 101.66 \, \text{J} \).
This equals the change in kinetic energy: \( \frac{1}{2}mv^2 - 0 \) (since starts from rest).
\( \frac{1}{2} \cdot 5 \cdot v^2 = 101.66 \).
\( 2.5v^2 = 101.66 \), so \( v^2 = 40.664 \), and \( v \approx \sqrt{40.664} \approx 6.38 \, \text{m/s} \).
The speed of the block after 2.60 m is approximately \( 6.38 \, \text{m/s} \).
1.5
a. The work done by the engine is calculated using \( W = F \cdot d \cdot \cos(\theta) \), where \( F = 200 \, \text{N} \), \( d = 4.8 \, \text{m} \), and \( \theta = 0^\circ \) (since the force is along the displacement).
\( W = 200 \cdot 4.8 \cdot \cos(0^\circ) = 200 \cdot 4.8 \cdot 1 = 960 \, \text{J} \).
The work done by the engine is \( 960 \, \text{J} \).
b. Using the work-energy principle: Net work = change in kinetic energy.
Net work \( W_{\text{net}} = W_{\text{engine}} + W_{\text{gravity}} + W_{\text{friction}} \).
- Work done by engine: \( 960 \, \text{J} \).
- Work done by gravity: \( W_{\text{gravity}} = mg \cdot d \cdot \sin(\theta) \) (positive because it acts down the incline).
\( m = 720 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( d = 4.8 \, \text{m} \), \( \theta = 42^\circ \).
\( W_{\text{gravity}} = 720 \cdot 9.8 \cdot 4.8 \cdot \sin(42^\circ) \).
\( \sin(42^\circ) \approx 0.6691 \), so \( W_{\text{gravity}} \approx 720 \cdot 9.8 \cdot 4.8 \cdot 0.6691 \approx 22740.5 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -f \cdot d = -23 \cdot 4.8 = -110.4 \, \text{J} \).
Net work: \( W_{\text{net}} = 960 + 22740.5 - 110.4 = 23590.1 \, \text{J} \).
This equals \( \frac{1}{2}mv^2 \):
\( \frac{1}{2} \cdot 720 \cdot v^2 = 23590.1 \).
\( 360v^2 = 23590.1 \), so \( v^2 = 65.528 \), and \( v \approx \sqrt{65.528} \approx 8.09 \, \text{m/s} \).
The velocity of the car after moving 4.8 m is approximately \( 8.09 \, \text{m/s} \).
a. The work done by gravity is calculated using the formula:
\( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the component of the force in the direction of displacement.
The gravitational force component parallel to the incline is \( mg \sin(\theta) \).
Here, \( m = 60 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( \theta = 40^\circ \), and \( d = 3 \, \text{m} \).
So, \( W = (60 \cdot 9.8 \cdot \sin(40^\circ)) \cdot 3 \).
\( \sin(40^\circ) \approx 0.6428 \), so:
\( W = (60 \cdot 9.8 \cdot 0.6428) \cdot 3 \approx (378.7) \cdot 3 \approx 1136.1 \, \text{J} \).
The work done by gravity is approximately \( 1136 \, \text{J} \).
b. The normal force is perpendicular to the displacement (which is along the incline). Therefore, the angle between the normal force and displacement is \( 90^\circ \), and \( \cos(90^\circ) = 0 \).
So, the work done by the normal force is \( 0 \, \text{J} \).
1.2
a. The work done by gravity is zero because the displacement is horizontal and gravity acts vertically. The angle between the force of gravity and displacement is \( 90^\circ \), so \( W = F \cdot d \cdot \cos(90^\circ) = 0 \).
The work done by gravity is \( 0 \, \text{J} \).
b. The work done by the applied force is calculated using \( W = F \cdot d \cdot \cos(\theta) \), where \( F = 60 \, \text{N} \), \( d = 3.25 \, \text{m} \), and \( \theta = 30^\circ \).
\( W = 60 \cdot 3.25 \cdot \cos(30^\circ) \).
\( \cos(30^\circ) \approx 0.866 \), so:
\( W = 60 \cdot 3.25 \cdot 0.866 \approx 171.9 \, \text{J} \).
The work done by the applied force is approximately \( 172 \, \text{J} \).
1.3
a. Free body diagram:
- Gravitational force (\( mg \)) acting vertically downward.
- Normal force (\( N \)) acting perpendicular to the incline.
- Applied force (\( F_{\text{applied}} = 8000 \, \text{N} \)) acting parallel to the incline upward.
- Frictional force (\( f = 20 \, \text{N} \)) acting parallel to the incline downward (opposite to motion).
b. The net work done on the car is the sum of the work done by each force.
- Work done by gravity: \( W_{\text{gravity}} = -mg \cdot d \cdot \sin(\theta) \) (negative because it opposes motion up the incline).
\( m = 1200 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( d = 3 \, \text{m} \), \( \theta = 30^\circ \).
\( W_{\text{gravity}} = -1200 \cdot 9.8 \cdot 3 \cdot \sin(30^\circ) = -1200 \cdot 9.8 \cdot 3 \cdot 0.5 = -17640 \, \text{J} \).
- Work done by normal force: \( 0 \, \text{J} \) (perpendicular to displacement).
- Work done by applied force: \( W_{\text{applied}} = 8000 \cdot 3 = 24000 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -20 \cdot 3 = -60 \, \text{J} \).
Net work: \( W_{\text{net}} = -17640 + 0 + 24000 - 60 = 6300 \, \text{J} \).
The net work done on the car is \( 6300 \, \text{J} \).
1.4
Using the work-energy principle: Net work done = change in kinetic energy.
Net work \( W_{\text{net}} = W_{\text{applied}} + W_{\text{friction}} \).
- Work done by applied force: \( W_{\text{applied}} = F \cdot d \cdot \cos(\theta) = 50 \cdot 2.60 \cdot \cos(30^\circ) \).
\( \cos(30^\circ) \approx 0.866 \), so \( W_{\text{applied}} = 50 \cdot 2.60 \cdot 0.866 \approx 112.58 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -f \cdot d = -4.2 \cdot 2.60 = -10.92 \, \text{J} \).
Net work: \( W_{\text{net}} = 112.58 - 10.92 = 101.66 \, \text{J} \).
This equals the change in kinetic energy: \( \frac{1}{2}mv^2 - 0 \) (since starts from rest).
\( \frac{1}{2} \cdot 5 \cdot v^2 = 101.66 \).
\( 2.5v^2 = 101.66 \), so \( v^2 = 40.664 \), and \( v \approx \sqrt{40.664} \approx 6.38 \, \text{m/s} \).
The speed of the block after 2.60 m is approximately \( 6.38 \, \text{m/s} \).
1.5
a. The work done by the engine is calculated using \( W = F \cdot d \cdot \cos(\theta) \), where \( F = 200 \, \text{N} \), \( d = 4.8 \, \text{m} \), and \( \theta = 0^\circ \) (since the force is along the displacement).
\( W = 200 \cdot 4.8 \cdot \cos(0^\circ) = 200 \cdot 4.8 \cdot 1 = 960 \, \text{J} \).
The work done by the engine is \( 960 \, \text{J} \).
b. Using the work-energy principle: Net work = change in kinetic energy.
Net work \( W_{\text{net}} = W_{\text{engine}} + W_{\text{gravity}} + W_{\text{friction}} \).
- Work done by engine: \( 960 \, \text{J} \).
- Work done by gravity: \( W_{\text{gravity}} = mg \cdot d \cdot \sin(\theta) \) (positive because it acts down the incline).
\( m = 720 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( d = 4.8 \, \text{m} \), \( \theta = 42^\circ \).
\( W_{\text{gravity}} = 720 \cdot 9.8 \cdot 4.8 \cdot \sin(42^\circ) \).
\( \sin(42^\circ) \approx 0.6691 \), so \( W_{\text{gravity}} \approx 720 \cdot 9.8 \cdot 4.8 \cdot 0.6691 \approx 22740.5 \, \text{J} \).
- Work done by friction: \( W_{\text{friction}} = -f \cdot d = -23 \cdot 4.8 = -110.4 \, \text{J} \).
Net work: \( W_{\text{net}} = 960 + 22740.5 - 110.4 = 23590.1 \, \text{J} \).
This equals \( \frac{1}{2}mv^2 \):
\( \frac{1}{2} \cdot 720 \cdot v^2 = 23590.1 \).
\( 360v^2 = 23590.1 \), so \( v^2 = 65.528 \), and \( v \approx \sqrt{65.528} \approx 8.09 \, \text{m/s} \).
The velocity of the car after moving 4.8 m is approximately \( 8.09 \, \text{m/s} \).
Parent Tip: Review the logic above to help your child master the concept of work force and power worksheet.
