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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The uniform crate has a mass of 150 kg. If the coefficient of static friction between the crate and the floor is us = 0.2, determine the smallest mass of the man so he can move the crate. The coefficient of static friction between his shoes and the floor is us = 0.45.
Assume the man exerts only a horizontal force on the crate
2.4 m
1.6 m
-1.2 m-
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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300 mm
A) 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…
Chapter 5 Solutions
COLLEGE PHYSICS
Ch. 5 - Prob. 1QAPCh. 5 - Prob. 2QAPCh. 5 - Prob. 3QAPCh. 5 - Prob. 4QAPCh. 5 - Prob. 5QAPCh. 5 - Prob. 6QAPCh. 5 - Prob. 7QAPCh. 5 - Prob. 8QAPCh. 5 - Prob. 9QAPCh. 5 - Prob. 10QAP
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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_forward1. Consider the forces acting on these three changes and list in order of the magnitude of the force starting with the smallest. A --- r B -4 +q +q Find FA Find Fa Find Fe List in order of the magnitudearrow_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_forwardX Figure 5-29 Three astronauts, propelled by jet backpacks, push and guide a 120 kg asteroid toward a processing dock, exerting the forces shown in Figure 5-29, with F1 = 32 N, F2 = 54 N, F3 = 41 N, 01 = 30°, and 03 = 60°. What is the asteroid's acceleration in unit-vector notation?arrow_forwardProblem 3 A 4-kg block rests on a horizontal surface. The block is pulled by 20 N force acting 25° above the horizontal but not enough to move the block. Calculate the following: (a) the friction force between the surface and the block. (b) the normal force on the block. (c) the coefficient of static friction.arrow_forward
- 1 Figure 5-19 gives the free-body diagram for four situations in which an object is pulled by several forces across a frictionless floor, as seen from overhead. In which situations does the acceleration a of the object have (a) an x component and (b) a y component? (c) In each situation, give the direction of a by naming either a quadrant or a direction along an axis. (Don't reach for the calculator because this can be answered with a few mental calculat s.) A7N 46N 46 N 3 N 3 N 2NT 3 N 2 N 3 N 5 N 2 N 5 N 2NV 4N 5 N v4 N ,3N 4 N ,4N 5 N V4 N (1) (2) (3) (4)arrow_forwardTwo 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_forward
- Question 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_forwardFlying Circus of Physics In the figure, a climber leans out against a vertical ice wall that has negligible friction. Distance a is 0.945 m and distance L is 1.95 m. His center of mass is distance d = 0.92 m from the feet-ground contact point. If he is on the verge of sliding, what is the coefficient of static friction between feet and ground? com —а — Hs = Number i Units %3Darrow_forwardQuestion 7 A 81.17 kg crate is on a frictionless surface inclined at an angle of 0 = 28.64°. A constant external force, P, is applied horizontally to the crate such that it moves a distance of 3.00 m up the incline at constant speed. What is the magnitude of the force P? (740) %3D P. 3.0 m A) O1303.27 N B) O381.27 N C) 0434.42 N D) O14.09 N E) 698.14 N Review Laterarrow_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