College Physics I PHYS 1301.001 - Homework 9 and Exam 2 Notes
College Physics I PHYS 1301.001
- Homework 9:
- Due on Mastering Physics, Monday, April 7.
- Exam 2:
- Review/Discussion Sessions (Dr. Khan):
- SCI 3.265
- Wed 11 am to 12 pm
- Thu 1 pm to 2 pm
Supplementary Instruction
- Tahir Ali:
- Tuesday at 4-5:15 (MC 3.606B)
- Thursday at 4-5:15 (MC 3.606B)
- Physics! Help! Online Problem Session (OPS) on Teams
- Crosby Pineda, Crosby.Pineda@utdallas.edu
- Monday 1 pm - 2 pm SCI 3.253
- 2 pm- 3 pm SCI 3.265
- Muhammad Khalid, Muhammad.Khalid@utdallas.edu
- Tuesday and Thursday 12 pm - 1 pm, SCI 2.179
- Meghraj Magadi Shivalingaiah, Meghraj.MagadiShivalingaiah@utdallas.edu
- Friday 2 pm - 3 pm, SCI 2.112
- Anusha Srivastava, Anusha.Srivastava@utdallas.edu
- Monday 2 pm - 3 pm, SCI B.139
- Mahammed Patel, mxp230007@utdallas.edu
- Tuesday 10.30 am to 11.30 am and 1.00 pm to pm, SCI 3.253
- Rittik Patra, rxp200009@utdallas.edu
- Thursday 2:30 pm-3:30 pm, SCI 2.112
Work and Energy (Chapter 7)
- Overview of Energy
- Work
- Work and Kinetic Energy
- Work Done with a Varying Force
- Potential Energy
- Conservation of Energy
- Conservative and Non-Conservative Forces
- Power
Work and Energy with Varying Forces
- Many forces are not constant.
- Suppose a particle moves along the x-axis from x1 to x2.
- The total work done by the force is represented by the area under the curve between the initial and final positions on a graph of force as a function of position.
Stretching a Spring
- Force due to elongation/compression of a spring follows Hooke’s law:
- This is a prime example of a varying force.
- Hooke's Law: F = -kx where:
- F is the force exerted by the spring.
- k is the spring constant (a measure of the stiffness of the spring).
- x is the displacement from the equilibrium position.
- The work done by stretching/compressing a spring is equal to the area of the shaded triangle:
Work Done on a Spring Scale
- Problem:
- A woman weighing 600 N steps onto a bathroom scale containing a heavy spring. The spring compresses by 1.0 cm under her weight.
- Find:
- The force constant of the spring.
- The total work done on it during the compression.
Solution for Spring Scale Problem
- Choose +x to be upwards (elongation).
- x = -0.01 m when F_x = -600 N
- Use Hooke’s Law to find k.
- F = -kx
- -600 = -k(0.01)
- k = 6.0 \times 10^4 N/m
- W = 3.0 Nm = 3.0 J
Potential Energy
- Potential Energy: Energy associated with the position of a system rather than its motion
- Gravitational Potential Energy:
- When a particle is in the gravitational field of the earth, there is a gravitational potential energy associated with the particle.
- As the basketball descends, gravitational potential energy is converted to kinetic energy, and the basketball’s speed increases.
Gravitational Potential Energy
- The quantity, product of the weight mg and the height y above the origin of coordinates, is gravitational potential energy.
- We can describe the work done from y1 to y2.
- The negative sign is important:
- When the body moves up, y increases, W{grav} is negative, and gravitational potential energy increases (\Delta U{grav} > 0).
- When the body moves down, y decreases, W{grav} is positive, and gravitational potential energy decreases (\Delta U{grav} < 0).
Conservation of Mechanical Energy (gravitational forces only)
- The total mechanical energy of a system is the sum of its kinetic energy and potential energy.
- A quantity that always has the same value is called a conserved quantity.
- When only the force of gravity does work on a system, the total mechanical energy of that system is conserved.
- This is an example of the conservation of mechanical energy.
Example 7.7 Conservation of Mechanical Energy (gravitational forces only)
- You throw a 0.145-kg baseball straight up, giving it an initial velocity of magnitude 20.0 m/s. Find how high it goes, ignoring air resistance.
Solution for Baseball Example
- After a baseball leaves your hand, mechanical energy E = K + U_{grav} is conserved.
- Given:
- Mass, m = 0.145 kg
- Initial velocity, v_1 = 20.0 m/s
- Energy at y_1 (initial position):
- y_1 = 0
- E = K1 + U{grav, 1}
- Energy at y_2 (highest point):
- v_2 = 0
- E = K2 + U{grav, 2}
- Conservation of energy:
- K1 + U{grav, 1} = K2 + U{grav, 2}
- \frac{1}{2}mv1^2 + mgy1 = \frac{1}{2}mv2^2 + mgy2
- Since y1 = 0 and v2 = 0:
- Solving for y_2:
- y2 = \frac{v1^2}{2g} = \frac{(20.0 \text{ m/s})^2}{2(9.80 \text{ m/s}^2)} = 20.4 \text{ m}
Work and Energy Along a Curved Path
- We can use the same expression for gravitational potential energy whether the body’s path is curved or straight.
- The total work done by the gravitational force depends only on the difference in height between the two points of the path.
- Unaffected by any horizontal motion that may occur.
- mg\Delta s \cos{\beta} = -mg\Delta y
- We can use the same expression for gravitational potential energy whether the body’s path is curved or straight.