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:
  • In class, April 17.
• 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
  • Friday 3 pm
• 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 $x<em>1$ to $x</em>2$.
• 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:
  • $W = \frac{1}{2}kx^2$

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.
  • $U_{grav} = mgy$
• We can describe the work done from $y<em>1$ to $y</em>2$.
  • $W<em>{grav} = mgy</em>1 - mgy_2$
• The negative sign is important:
  • When the body moves up, $y$ increases, $W<em>{grav}$ is negative, and gravitational potential energy increases (\Delta U{grav} > 0).
  • When the body moves down, $y$ decreases, $W<em>{grav}$ 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.
  • $E = K + U$
• 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.
  • $K<em>1 + U</em>1 = K<em>2 + U</em>2$
• 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):
  • $y_1 = 0$
  • $E = K<em>1 + U</em>{grav, 1}$
• Energy at $y_2$ (highest point):
  • $v_2 = 0$
  • $E = K<em>2 + U</em>{grav, 2}$
• Conservation of energy:
  • $K<em>1 + U</em>{grav, 1} = K<em>2 + U</em>{grav, 2}$
  • $\frac{1}{2}mv<em>1^2 + mgy</em>1 = \frac{1}{2}mv<em>2^2 + mgy</em>2$
  • Since $y<em>1 = 0$ and $v</em>2 = 0$:
    • $\frac{1}{2}mv<em>1^2 = mgy</em>2$
• Solving for $y_2$:
  • $y<em>2 = \frac{v</em>1^2}{2g} = \frac{(20.0 \text{ m/s})^2}{2(9.80 \text{ m/s}^2)} = 20.4 \text{ m}$

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.
  • $W<em>{grav} = mgy</em>1 - mgy_2$
• 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.