The Following Questions Ask You To Solve Problems Inv 215103
The Following Questions Ask You To Solve Problems Involving Topics We
The following questions ask you to solve problems involving topics we've covered this week. Work each problem on paper, and post a scanned image of your work for each question, along with a brief description of how you solved the problem. Your scanned work should include a step-by-step solution to the problem. Upload your work to the same W4 Assignment 2 Lab Dropbox.
Paper For Above instruction
Question 1
A person pulls a crate with 120 N of force against 100 N of friction a distance of 5 meters. Sketch a free body diagram showing the forces acting on the crate and find the energy the person expends pulling the crate, the amount of energy converted to heat via friction, and the increase in the kinetic energy of the crate.
Question 2
A box of mass 4 kg is resting on a table 1.5 m above the floor. There is also a ledge located 5 m above the floor. First, provide a sketch of this problem. Then, find the gravitational potential energy of the box relative to the floor, and the gravitational potential energy of the box relative to the ledge.
Question 3
An 80 kg man climbs a 50 m-high hill in 20 minutes. What is the average power in watts he expends climbing the hill?
Question 4
A 40,000 kg railroad car initially traveling at 10 m/s collides inelastically with a 20,000 kg railroad car initially at rest. The cars stick together. What is their final speed?
Question 5
A roller coaster is initially at a height of 40 m above the ground and has an initial velocity of 15 m/s. Using conservation of energy, find the velocity of the roller coaster at a height of 5 m above the ground.
Paper For Above instruction
Question 1: Energy Analysis of Pulling a Crate
The problem involves analyzing the forces and energy expenditure when pulling a crate against friction over a set distance. The free body diagram includes the pulling force, the frictional force, and the normal force and gravity acting on the crate. The work done by the person is calculated by multiplying the pulling force by the distance moved in the direction of the force. Since the crate is pulled a distance of 5 meters with a force of 120 N, the work expended by the person is:
\[ W_{person} = F_{pull} \times d = 120\, \text{N} \times 5\, \text{m} = 600\, \text{J} \]
The energy converted to heat due to friction is the work done against the frictional force:
\[ W_{friction} = F_{friction} \times d = 100\, \text{N} \times 5\, \text{m} = 500\, \text{J} \]
The net work done on the crate results in an increase in its kinetic energy:
\[ \Delta KE = W_{person} - W_{friction} = 600\, \text{J} - 500\, \text{J} = 100\, \text{J} \]
Thus, the crate gains 100 Joules of kinetic energy.
Question 2: Gravitational Potential Energy of a Box
The gravitational potential energy (GPE) is given by:
\[ U = mgh \]
where \( m = 4\, \text{kg} \).
- Relative to the floor (at height \( h = 1.5\, \text{m} \)):
\[ U_{floor} = 4\, \text{kg} \times 9.8\, \text{m/s}^2 \times 1.5\, \text{m} = 58.8\, \text{J} \]
- Relative to the ledge (at height \( h = 5\, \text{m} \)):
\[ U_{ledge} = 4\, \text{kg} \times 9.8\, \text{m/s}^2 \times 5\, \text{m} = 196\, \text{J} \]
The potential energy relative to the floor is 58.8 J, and relative to the ledge, it is 196 J.
Question 3: Power Expended Climbing a Hill
The work done to elevate the man is the change in gravitational potential energy:
\[ W = mgh = 80\, \text{kg} \times 9.8\, \text{m/s}^2 \times 50\, \text{m} = 39,200\, \text{J} \]
Time taken:
\[ t = 20\, \text{minutes} = 1200\, \text{s} \]
The average power \( P \):
\[ P = \frac{W}{t} = \frac{39,200\, \text{J}}{1200\, \text{s}} \approx 32.67\, \text{W} \]
The man expends approximately 32.67 watts on average.
Question 4: Final Speed of Colliding Railroad Cars
Using conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]
where \( m_1 = 40000\, \text{kg} \), \( v_1 = 10\, \text{m/s} \), \( m_2 = 20000\, \text{kg} \), \( v_2 = 0 \):
\[ (40000\, \text{kg} \times 10\, \text{m/s}) + (20000\, \text{kg} \times 0) = (60000\, \text{kg}) \times v_f \]
\[ 400,000\, \text{kg m/s} = 60000\, \text{kg} \times v_f \]
\[ v_f = \frac{400,000}{60,000} \approx 6.67\, \text{m/s} \]
The final velocity of the combined cars is approximately 6.67 m/s.
Question 5: Velocity of Roller Coaster at 5 m Height
Initial mechanical energy (at 40 m):
\[ E_{initial} = PE + KE = mgh + \frac{1}{2} mv^2 \]
\[ E_{initial} = 1000\, \text{kg} \times 9.8\, \text{m/s}^2 \times 40\, \text{m} + \frac{1}{2} \times 1000\, \text{kg} \times (15\, \text{m/s})^2 \]
\[ E_{initial} = 392,000\, \text{J} + 112,500\, \text{J} = 504,500\, \text{J} \]
At 5 m height:
\[ PE_{final} = 1000\, \text{kg} \times 9.8\, \text{m/s}^2 \times 5\, \text{m} = 49,000\, \text{J} \]
Using conservation of energy:
\[ E_{initial} = PE_{final} + KE_{final} \]
\[ 504,500\, \text{J} = 49,000\, \text{J} + KE_{final} \]
\[ KE_{final} = 455,500\, \text{J} \]
Solve for velocity:
\[ KE = \frac{1}{2} mv^2 \]
\[ v = \sqrt{\frac{2 KE}{m}} = \sqrt{\frac{2 \times 455,500\, \text{J}}{1000\, \text{kg}}} \approx \sqrt{911} \approx 30.2\, \text{m/s} \]
The velocity at 5 m height is approximately 30.2 m/s.
References
- Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
- Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (9th ed.). Cengage Learning.
- Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed.). Pearson.
- Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers. W. H. Freeman.
- Knight, R. D. (2013). Physics for Scientists and Engineers. Pearson.
- Young, H. D., & Freedman, R. A. (2019). University Physics with Modern Physics. Pearson.
- Taylor, J. R. (2005). Classical Mechanics. University Science Books.
- Thornton, S. T., & Marion, J. B. (2003). Classical Dynamics. Brooks Cole.
- Feynman, R. P., Leighton, R. B., & Sands, M. (2010). The Feynman Lectures on Physics. Addison-Wesley.
- Ohanian, H. C., & Ruffini, R. (2013). Gravitation and Spacetime. Cambridge University Press.