
Concept explainers
(a)
To Calculate: The maximum and minimum values of applied force for which the block does not slip.
(a)

Answer to Problem 63P
Explanation of Solution
Given data:
Mass of the block,
Mass of the wedge,
Coefficient of static friction,
Angle of wedge,
Formula Used:
Newton’s second law of motion:
Where, m is the mass and a is the acceleration.
Calculation:
Free-body diagram:
Maximum force,
Minimum force,
Where,
For the block,
Substitute (3) in (1) to find the minimum acceleration:
Substitute the values and solve:
For maximum force, reverse the direction of f
Substitute the values and solve:
Conclusion:
The maximum and minimum values of applied force for which the block does not slip are 84 N and 1.6 N respectively.
(b)
To Calculate: The maximum and minimum values of applied force for which the block does not slip.
(b)

Answer to Problem 63P
Explanation of Solution
Given data:
Mass of the block,
Mass of the wedge,
Coefficient of static friction,
Angle of wedge,
Formula Used:
From previous part:
Calculation:
Substitute the values and solve for minimum force:
For maximum force, reverse the direction of f
Substitute the values and solve:
Conclusion:
The maximum and minimum values of applied force for which the block does not slip are 37.5 N and 5.8 N respectively.
Want to see more full solutions like this?
Chapter 5 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- Find the current in 5.00 and 7.00 Ω resistors. Please explain all reasoningarrow_forwardFind the amplitude, wavelength, period, and the speed of the wave.arrow_forwardA long solenoid of length 6.70 × 10-2 m and cross-sectional area 5.0 × 10-5 m² contains 6500 turns per meter of length. Determine the emf induced in the solenoid when the current in the solenoid changes from 0 to 1.5 A during the time interval from 0 to 0.20 s. Number Unitsarrow_forward
- A coat hanger of mass m = 0.255 kg oscillates on a peg as a physical pendulum as shown in the figure below. The distance from the pivot to the center of mass of the coat hanger is d = 18.0 cm and the period of the motion is T = 1.37 s. Find the moment of inertia of the coat hanger about the pivot.arrow_forwardReview Conceptual Example 3 and the drawing as an aid in solving this problem. A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 3.9 m/s perpendicular to a 0.49-T magnetic field. The resistance of th rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.4 m. A 1.1-Q resistor is attached between the tops of the tracks. (a) What is the mass of the rod? (b) Find the change in the gravitational potentia energy that occurs in a time of 0.26 s. (c) Find the electrical energy dissipated in the resistor in 0.26 s.arrow_forwardA camera lens used for taking close-up photographs has a focal length of 21.5 mm. The farthest it can be placed from the film is 34.0 mm. (a) What is the closest object (in mm) that can be photographed? 58.5 mm (b) What is the magnification of this closest object? 0.581 × ×arrow_forward
- Given two particles with Q = 4.40-µC charges as shown in the figure below and a particle with charge q = 1.40 ✕ 10−18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞.) Three positively charged particles lie along the x-axis of the x y coordinate plane.Charge q is at the origin.Charge Q is at (0.800 m, 0).Another charge Q is at (−0.800 m, 0).(a)What is the net force (in N) exerted by the two 4.40-µC charges on the charge q? (Enter the magnitude.) N(b)What is the electric field (in N/C) at the origin due to the two 4.40-µC particles? (Enter the magnitude.) N/C(c)What is the electrical potential (in kV) at the origin due to the two 4.40-µC particles? kV(d)What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 4.40-µC particles?arrow_forward(a) Where does an object need to be placed relative to a microscope in cm from the objective lens for its 0.500 cm focal length objective to produce a magnification of -25? (Give your answer to at least three decimal places.) 0.42 × cm (b) Where should the 5.00 cm focal length eyepiece be placed in cm behind the objective lens to produce a further fourfold (4.00) magnification? 15 × cmarrow_forwardIn a LASIK vision correction, the power of a patient's eye is increased by 3.10 D. Assuming this produces normal close vision, what was the patient's near point in m before the procedure? (The power for normal close vision is 54.0 D, and the lens-to-retina distance is 2.00 cm.) 0.98 x marrow_forward
- Don't use ai to answer I will report you answerarrow_forwardA shopper standing 2.00 m from a convex security mirror sees his image with a magnification of 0.200. (Explicitly show on paper how you follow the steps in the Problem-Solving Strategy for mirrors found on page 1020. Your instructor may ask you to turn in this work.) (a) Where is his image (in m)? (Use the correct sign.) -0.4 m in front of the mirror ▾ (b) What is the focal length (in m) of the mirror? -0.5 m (c) What is its radius of curvature (in m)? -1.0 marrow_forwardAn amoeba is 0.309 cm away from the 0.304 cm focal length objective lens of a microscope.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





