Tag Archives: College Physics Solutions
Chapter 7: Work, Energy, and Energy Resources
Work: The Scientific Definition
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Kinetic Energy and the Work-Energy Theorem
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Gravitational Potential Energy
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Conservative Forces and Potential Energy
Problem 22
Problem 23
Nonconservative Forces
Problem 24
Problem 25
Conservation of Energy
Problem 26
Problem 27
Problem 28
Problem 29
Power
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Work, Energy, and Power in Humans
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
World Energy Use
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Chapter 6: Uniform Circular Motion and Gravitation
Rotation Angle and Angular Velocity
Centripetal Acceleration
Problem 22
Centripetal Force
Problem 31
Problem 32
Newton’s Universal Law of Gravitation
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Satellites and Kepler’s Laws: An Argument for Simplicity
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity
Friction
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Drag Forces
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Elasticity: Stress and Strain
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Chapter 4: Dynamics: Force and Newton’s Laws of Motion
Newton’s Second Law of Motion: Concept of a System
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Newton’s Third Law of Motion: Symmetry in Forces
Problem 15
Problem 16
Normal, Tension, and Other Example of Forces
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem Solving Strategies
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Further Applications of Newton’s Laws of Motion
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Further Applications of Newton’s Laws of Motion
Problem 52
Problem 53
Problem 54
Chapter 3: Two-Dimensional Kinematics
Vector Addition and Subtraction: Graphical Methods
Vector Addition and Subtraction: Analytical Methods
Projectile Motion
Addition of Velocities
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Physics
College Physics by Openstax Chapter 3 Problem 37
Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?
Solution:
Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.
College Physics by Openstax Chapter 3 Problem 36
The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.
Solution:
We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:
- The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
- The jumper is on level ground, and the motion started from the ground.
The formula for range is
\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}
Since we are already given the necessary details, we can now solve for the range.
\begin{align*} \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\ \text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
College Physics by Openstax Chapter 3 Problem 35
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Solution:
We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:
- The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
- The jumper is on level ground.
The formula for the range is
\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}
To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.
\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}
In this case, the final velocity will be the initial velocity of the jump.
\begin{align*} \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s} \end{align*}
So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.
\begin{align*} \text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\ \text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\ \text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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