
(a)
The extension of the spring for a mass of
(a)

Answer to Problem 65P
The extension of the spring for a mass of
Explanation of Solution
Write the expression for
Here,
Write the expression for velocity in terms of time period.
Here,
Write the expression for force from hooks law.
Here,
Use equation (II) and (III) in equation (I) and rearrange.
Write the expression for radius of the pluck’s motion.
Use equation (V) in equation (IV), to find
Conclusion:
Therefore, the extension of the spring for a mass of
(b)
The extension of the spring for the mass
(b)

Answer to Problem 65P
The extension of the spring for the mass
Explanation of Solution
Substitute
Conclusion:
Substitute
Therefore, the extension of the spring for the mass
(c)
The extension of the spring for the mass
(c)

Answer to Problem 65P
The extension of the spring for the mass
Explanation of Solution
From equation (VII).
Conclusion:
Substitute
Therefore, the extension of the spring for the mass
(d)
The extension of the spring for the mass
(d)

Answer to Problem 65P
The extension of the spring for the mass
Explanation of Solution
From equation (VII).
Conclusion:
Substitute
Therefore, the extension of the spring for the mass
(e)
The extension of the spring for the mass
(e)

Answer to Problem 65P
For the mass
Explanation of Solution
From equation (VII) the spring extension is given by
Conclusion:
Substitute
Therefore, For the mass
(f)
To explain the pattern of variation of
(f)

Answer to Problem 65P
The extension of the spring is directly proportional to the mass
Explanation of Solution
The extension of the spring is directly proportional to the mass
Conclusion:
Therefore, the extension of the spring is directly proportional to the mass
Want to see more full solutions like this?
Chapter 6 Solutions
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
- Min Min is hanging from her spring-arms off the edge of the level. Due to the spring like nature of her arms she is bouncing up and down in simple harmonic motion with a maximum displacement from equilibrium of 0.118 m. The spring constant of Min-Min’s arms is 9560. N/m and she has a mass of 87.5 kg. What is the period at which she oscillates? Find her maximum speed. Find her speed when she is located 5.00 cm from her equilibrium position.arrow_forward(a) What magnification in multiples is produced by a 0.150 cm focal length microscope objective that is 0.160 cm from the object being viewed? 15.9 (b) What is the overall magnification in multiples if an eyepiece that produces a magnification of 7.90x is used? 126 × ×arrow_forwardGravitational Potential Energyarrow_forward
- E = кедо Xo A continuous line of charge lies along the x axis, extending from x = +x to positive infinity. The line carries positive charge with a uniform linear charge density 10. (a) What is the magnitude of the electric field at the origin? (Use the following as necessary: 10, Xo, and ke.) (b) What is the direction of the electric field at the origin? O O O O O O G -y +z ○ -z +x -x +yarrow_forwardInclude free body diagramarrow_forward2 Spring 2025 -03 PITT Calculate the acceleration of a skier heading down a 10.0° slope, assuming the coefficient of cold coast at a constant velocity. You can neglect air resistance in both parts. friction for waxed wood on wet snow fly 0.1 (b) Find the angle of the slope down which this skier Given: 9 = ? 8=10° 4=0.1arrow_forward
- dry 5. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a c piston into a steel cylinder. What is the normal force between the piston and cyli=030 What force would she have to exert if the steel parts were oiled? k F = 306N 2 =0.03 (arrow_forwardInclude free body diagramarrow_forwardInclude free body diagramarrow_forward
- Test 2 МК 02 5. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a dry = 0.03 (15 pts) piston into a steel cylinder. What is the normal force between the piston and cylinder? What force would she have to exert if the steel parts were oiled? Mk Giren F = 306N MK-0.3 UK = 0.03 NF = ?arrow_forward2. A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 ke? a = 350 m/s 2arrow_forward2. A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? (10 pts) a = 3.50 m/s 2 distance 90 km/h = 3.50m/62 M = 245garrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





