COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 5, Problem 46QAP
To determine
The acceleration of the box.
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8-63. Determine the smallest force P that will cause
impending motion. The crate and wheel have a mass of
50 kg and 25 kg, respectively. The coefficient of static
friction between the crate and the ground is , = 0.2, and
between the wheel and the ground, = 0.5.
*8-64. Determine the smallest force P that will cause
impending motion. The crate and wheel have a mass of
50 kg and 25 kg. respectively. The coefficient of static
friction between the crate and the ground is , = 0.5, and
between the wheel and the ground μ = 0.3.
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Chapter 5 Solutions
COLLEGE PHYSICS
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- An automobile driver traveling down an 8% grade slams on his brakes and skids 30 m before hitting a parked car. A lawyer hires an expert who measures the coefficient of kinetic friction between the tires and road to be k = 0.45. Is the lawyer correct to accuse the driver of exceeding the 25-MPH speed limit? Explain.arrow_forwardA) A woman exerts a constant horizontal force on a large box. As a result, the box moves across a horizontal floor at a constant speed "v0". The constant horizontal force applied by the woman: has the same magnitude as the weight of the box. is greater than the weight of the box. has the same magnitude as the total force which resists the motion of the box. is greater than the total force which resists the motion of the box. is greater than either the weight of the box or the total force which resists its motion. B) If the woman in the previous question doubles the constant horizontal force that she exerts on the box to push it on the same horizontal floor, the box then moves: with a constant speed that is double the speed "v0" in the previous question. with a constant speed that is greater than the speed "v0" in the previous question, but not necessarily twice as great. for a while with a speed that is constant and greater than the speed "v0" in the previous question, then with a…arrow_forwardPRINTER VERSION 4 ВАСК NEXT Z Your answer is partially correct. Try again. A 2.70 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 5.98 N and a vertical force P are then applied to the block (see the figure). The coefficients of friction for the block and surface are us = 0.4 and u = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P is (a)10.0 N and (b)14.0 N. (The upward pull is insufficient to move the block vertically.) m em em (a) Number 5.9 Units em lem (b) Number T7.37 Units lem Click if you would like to Show Work for this question: Open Show Work lem plem SHOW HINT blem LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE blem e here to search 7:12 PM ENG 4/4/2021 13) 17 pause break eno Pgup prt sc delete hon insert 115Pgdn %23 3 4 5. backspo 远arrow_forward
- Part 2) A block with a mass of m = 0.410 kg is placed on a plane that can be tilted, as shown below. The coefficient of static friction between the block and the plane is 4, = 0.350, while the coefficient of kinetic friction between the block and the plane is Hk = 0.270. The angle between the inclined plane and the block is slowly increased. No additional force is applied to the block. m At what angle will the block start to slide down the plane? Part 3) If the plane is at the angle you calculated in Part 2, what will be the acceleration of the block down the plane? a = m/s? down the planearrow_forward3arrow_forwardA 3.5 kg block is pushed along a horizontal floor by a force F of magnitude 15 N at an angle q = 40 with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of a) the frictional force on the block from the floor and b) the block's acceleration.arrow_forward
- Two blocks are connected by a string, as shown in the figure (Figure 1). The smooth inclined surface makes an angle of 35° with the horizontal, and the block on the incline has a mass of 5.7 kg. The mass of the hanging block is m = 3.1 kg. Part A For the steps and strategies involved in solving a similar problem, you may view the following Example 6-13 video: Find the direction of the hanging block's acceleration. REASONING AND STRATEGY O Upward We will use Newton's second lw to ink the forces. maes, and accelerations F. - ma, F,-ma, O Downward Add up all forces in each direction that act on each objecer, using the free-body diagram as a guide Find the acceleration components, and then apply Newson's second law. N4 Submit Request Answer Er. Tma E. , -T-ma Part B Er, N-, -0 Find the magnitude of the hanging block's acceleration. Express your answer in meters per second squared. να ΑΣφ. a = m/s? Submit Request Answer Figure Provide Feedback 5.7 kg m 35°arrow_forwardm2 m1 Question 6/13 In the figure below, if m,-7m, and the stationary system is just about to move. What is the value of the coefficient of static friction? 1. 00.20 2. O0.17 3. O0.14 4. O0.12 5. 00.10 Next Activate Windaws 50PM 90'F Haze Type here to searcharrow_forwardQuestion 5 A mug rests on an inclined surface, as shown in (Figure 1), 0 = 21° . Part A For the steps and strategies involved in solving a similar problem, you may view the following Example 6-3 video: What is the magnitude of the frictional force exerted on the mug? SOLUTION Express your answer using appropriate units. To solve for the coefficient of static friction, we apply Newton's second law along the x axis: EF, = mp μΑ mgsin 6 - fama = 0 esin mg sin e- ,N Value gcos a Submit Request Answer Part B What is the minimum coefficient of static friction required to keep the mug from sliding? Figure 1 of 1 Submit Request Answer 0.27 kg Provide Feedbackarrow_forward
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Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY