Physics 1 2 Hour Test 2 ✓ Solved

physics1 2hourstest2

Reduce the assignment questions to concise calculations. Answer each in order, covering all specified problems including work, energy, motion, oscillations, collision analysis, thermal energy, heat transfer, and radiation. Provide full detailed solutions, utilizing relevant formulas and constants, and include credible references. Conclude with a references section in HTML format.

Paper For Above Instructions

Problem 1: Find the work done by a spring extended from 2 cm to 10 cm

Given:

Spring constant, k = 100 N/m

Initial extension, x₁ = 2 cm = 0.02 m

Final extension, x₂ = 10 cm = 0.10 m

Work done, W = ?

The work done in extending or compressing a spring is given by:

W = 1/2 k (x₂² - x₁²)

Calculating:

W = 0.5 100 [(0.10)² - (0.02)²]

= 50 * (0.01 - 0.0004)

= 50 * 0.0096

= 0.48 Joules

Problem 2: Find the work done on a 2 kg mass under a specified force

Suppose the force experienced is known or given as a graph; for simplicity, assume a force F applied over distance d. If the force varies, integrate F along the path. For example, with a uniform force F over distance d:

Work, W = F * d

If F is variable, W = ∫ F dx over the limits.

Without specific force data, a generic formula applies:

W = F * d

Problem 3: Compute the final velocity of the 2 kg mass after work application

Using work-energy theorem:

W = Δ KE = (1/2) m v_f² - (1/2) m v_i²

Assuming initial velocity v_i = 0, then:

v_f = √(2 W / m)

Substitute the work obtained from Problem 2, or a similar value, to calculate v_f.

Problem 4: Mechanical power exerted by a student lifting a barbell

Given:

Mass, m = 50 kg

Number of lifts = 20

Total time = 60 seconds

Height per lift, h = 0.75 m

Total work = m g h * number of lifts

Total work = 50 9.8 0.75 20 = 50 9.8 15 = 50 147 = 7350 Joules

Power, P = Work / Time = 7350 / 60 ≈ 122.5 Watts

Problem 5: Energy used considering muscle efficiency

Efficiency, η = 20% = 0.20

Energy input = Work / η = 7350 / 0.20 = 36,750 Joules

This is the food energy consumed during the activity.

Problem 6: Work-energy equation for rocket engine

Work done by rocket engine when ascending from Earth's surface to altitude h at speed v:

W = KE_final + PE_g_final - KE_initial - PE_g_initial

Since starting from rest at Earth's surface, KE_initial = 0 and PE_g_initial can be taken as zero at surface. Final kinetic energy: (1/2) m v², final potential energy: m g h.

Equation:

W = (1/2) m v² + m g h

Problem 7: Period of oscillation for a mass-spring system

T = 2π √(m / k)

Given: m = 2 kg, k = 50 N/m

T = 2π √(2 / 50) = 2π √0.04 = 2π * 0.2 ≈ 1.2566 seconds

Problem 8: Sketching x-t and v-t graphs for SHM (descriptive)

In SHM with displacement x decreasing to zero and moving right, the x-t graph is a cosine wave starting at maximum, while the v-t graph is a sine wave starting at zero, indicating the mass moves toward equilibrium with increasing velocity.

Problem 9: Displacement x(t) for a mass in SHM

Initial conditions: at equilibrium, v = 5 m/s to the right.

Amplitude, A = unknown, angular frequency, ω = √(k/m) = √(100/4) = 5 rad/sec.

Initial velocity v_i = A ω.

Since initial position x(0) = 0 at equilibrium, x(t) = A sin(ω t) + x_0. For initial velocity 5 m/s:

v(0) = A ω cos(0) = A * 5 = 5, thus A = 1 m.

Equation: x(t) = 1 * sin(5 t)

Problem 10: Elastic stretch upon impact on trampoline

Using conservation of energy at maximum compression:

(1/2) m v² = (1/2) k x²

Solving for x:

x = √(m v² / k)

Substituting: m = 25 kg, v = 8 m/s, k = 1200 N/m

x = √(25 * 64 / 1200) = √(1600 / 1200) ≈ √1.33 ≈ 1.15 meters

Problem 11: Initial position and velocity from simple harmonic motion

Given: x(t) = 0.25 cos(3t + 0.25)

Initial position, at t=0: x(0) = 0.25 cos(0.25) ≈ 0.25 * 0.9689 ≈ 0.2422 m, to the right.

Initial velocity: v(t) = -A ω sin(ω t + φ)

At t=0: v(0) = -0.25 3 sin(0.25) ≈ -0.75 * 0.2474 ≈ -0.1856 m/s, moving left.

Problem 12: Center of mass coordinate and conservation equations for oblique collision

Center of mass coordinate:

X_cm = (m₁ x₁ + m₂ x₂) / (m₁ + m₂)

Law of conservation of momentum:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

Coefficient of restitution between colliding bodies:

e = (v₂' - v₁') / (v₁ - v₂)

Problem 14: Momentum and restitution equations for e=0.4 collision

Momentum conservation:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

Restitution:

0.4 = (v₂' - v₁') / (v₁ - v₂)

Problem 15: Equations for impact energy transfer to swing

Use conservation of momentum:

m_putty v_putty = (m_putty + m_bob) v_final

Energy at maximum height:

h = v_final² / (2 g)

Equation:

h = [(m_putty * v_putty) / (m_putty + m_bob)]² / (2 g)

Problem 16: Heat energy to convert water from 20°C to steam

Q = m (C ΔT + L_vaporization)

C_water = 4186 J/(kg°C)

ΔT = 100 - 20 = 80°C

L_vaporization ≈ 2.26 x 10^6 J/kg

Q = 1.5 (4186 80 + 2.26×10^6)

Q = 1.5 (334,880 + 2,260,000) ≈ 1.5 2,594,880 ≈ 3,892,320 Joules

Problem 17: Final temperature when steel added to water

Using heat balance:

m_steel C_steel (T_steel_initial - T) = m_water C_water (T - T_water_initial)

Equation:

20 450 (120 - T) = 5 4186 (T - 20)

Problem 18: Heat conduction rate through steel wall

Using Fourier's law:

Q/t = (k A ΔT) / d

Where:

k = 80 J/(s°C·m),

A = π r² = π (0.10 m)² ≈ 0.0314 m²,

ΔT = 250°C - 100°C = 150°C,

d = 0.01 m

Q/t = (80 0.0314 150) / 0.01 ≈ (80 0.0314 150) / 0.01 ≈ (80 * 4.71) / 0.01 ≈ 376.8 / 0.01 = 37,680 W

Problem 19: Wind chill temperature

Using the wind chill formula:

T_wc = 13.12 + 0.6215 T_a - 11.37 v^{0.16} + 0.3965 T_a v^{0.16}

Where T_a = air temperature (5°C), v = wind speed (approximate using h), h=10 for still air, and h=15 for windy.

Calculations:

T_wc ≈ 13.12 + 0.6215 5 - 11.37 15^{0.16} + 0.3965 5 15^{0.16}

Approximately, T_wc ≈ 13.12 + 3.11 - 11.37 1.64 + 0.3965 5 * 1.64

≈ 16.23 - 18.63 + 3.25 ≈ 0.87°C

Problem 20: Radiant cooling rate of object

Using Stefan-Boltzmann law:

Power radiated = ε σ A T^4, where T in Kelvin.

Calculate:

T_object = 30 + 273.15 = 303.15 K

T_room = 20 + 273.15 = 293.15 K

Power:

P = ε σ A (T_object^4 - T_room^4)

Constants: ε=0.82, σ=5.67×10^-8 W/(m²·K^4), A=3 m²

Calculate:

P = 0.82 5.67×10^-8 3 * [(303.15)^4 - (293.15)^4]

Compute the difference:

(303.15)^4 ≈ 8.514×10^{9}, (293.15)^4 ≈ 7.353×10^{9}

Difference ≈ 1.161×10^{9}

Finally:

P ≈ 0.82 5.67×10^-8 3 1.161×10^9 ≈ 0.82 5.67×10^-8 3 1.161×10^9 ≈ ≈ 0.82 (5.67×10^-8 3) * 1.161×10^9

≈ 0.82 1.701×10^-7 1.161×10^9 ≈ 0.82 * 197.4 ≈ 162.0 Watts

References

  • Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers.
  • Young, H. D., & Freedman, R. A. (2019). University Physics with Modern Physics.
  • Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers.
  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics.
  • Hibbeler, R. C. (2016). Engineering Mechanics: Dynamics.
  • Meriam, J. L., & Kraige, L. G. (2012). Engineering Mechanics: Dynamics and Statics.
  • Çengel, Y. A., & Boles, M. A. (2015). Thermodynamics: An Engineering Approach.
  • Feynman, R. P., Leighton, R. B., & Sands, M. (2011). The Feynman Lectures on Physics.
  • Bevington, P. R., & Robinson, D. K. (2003). Data Reduction and Error Analysis for the Physical Sciences.
  • Kreith, F., & Bohn, M. J. (2011). Principles of Heat Transfer.

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