COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 5, Problem 47QAP
To determine
The minimum magnitude of force required to start the crates in motion.
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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-
5) A 640-N hunter gets a rope around a 3200-N polar bear. They are initially stationary, 20 meters apart, on frictionless level ice. The hunter now pulls on the rope such that she and the polar bear slide closer together. This continues until they eventually meet at some point on the ice. What total distance will the polar bear have slid?
a) 1.0 m
b) 3.3 m
c) 10 m
d) 12 m
e) 17 m
elearn.squ.edu.om/mod/qui
NG SYSTEM (ACADEMIC)
al Physics I- Spring21
Time left 0:09:24
A dog and a sledge are on the frictionless ice of a frozen lake, 16.3 m
apart but connected by a rope of negligible mass. The dog exerts a certain
horizontal force (N) on the rope. If magnitudes of the sledge and the dog
accelerations are 0.3 m/s? and 0.1 m/s?, respectively. How far from the
dog's initial position (m) do they meet?
Answer:
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_forwardA block of ice (m = 15.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal force of 95.0 N for 1.54 s. a. Determine the magnitude of each force acting on the block of ice while you are pulling. b. With what speed is the ice moving after you are finished pulling? Repeat Problem 71, but this time you pull on the block at an angle of 20.0.arrow_forward8-63. 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. O O O 127 10 L C A 300 mmarrow_forward
- PROBLEM I The 260kg crate shown in the figure rests on a horizontal surface for which the coefficient of kinetic friction is 025. If the crate is subjected to a 400 N towing force as shown determine the velocity of the crate in 5 s starting from rest PROBLEM 2 PROBLEM 3 The 80 kg block A shown in Figure is released from rest If the masses of the puleys and the cord are neglected determine the speed of the 12 kg block B in 3 s m₁ P = 400 N 6 = 30° 30⁰ SH Datum Consider the masses my 20 kg and m, 18 kg in the system represented by the figure bekow, If the coefficient of friction is OJ and the incination angle is 30°, find the acceleration of the system and the tension in the cord joining two mossCS m₂arrow_forwardPROBLEMI The 260kg crate shown in the figure rests on a horizontal surface for which the coefficient of kinetic friction is 025. If the crate is subjected to a 400 N towing force as shown determine the velocity of the crate in 5 s starting from rest PROBLEM 2 PROBLEM 3 The 80 kg block A shown in Figure is released from rest If the masses of the puleys and the cord are neglected determine the speed of the 12 kg block B in 38 m₁ P = 400 N IIA 8 = 30° 30° Consider the masses m, 20 kg and m, 18 kg in the system represented by the figure below. If the coefficient of friction is Of and the inclination angle is 30°, find the acceleration of the system and the tension in the cord joining two masses Datum m₂arrow_forward3- The student pushes horizontally on a 37 kg box with a force of 115 N. The coefficient friction between the box and the floor is 0.37. What is the frictional force on the crate from the floor? a- 115 b- 230 c-110 d-121arrow_forward
- PRINTER 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_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 1 BACK NEXT Chapter 07, Problem 007 A 4.6 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 1.8 s? rt=0 0.5 s -1.0s 1.5 s 2.0 s- 0.2 0.4 0.6 0,8 x (m) Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER e to search 6:53 PM ENG 3/30/2021 ASUS 15 ghome 10end Pgup 12Pgdn pause bregk prt sc insert delete 23 3. 8. 9. backspacearrow_forward
- PROBLEM 2 • An elevator and its load have a total mass of 800kg.Find the tension in the supporting cable when the elevator or lift originally moving downwards at 10m/s is brought to rest with a constant acceleration in a distance 25m Answer T-9600N.Since T> W the system is retarding.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_forward8) Two blocks of masses ml and m2 are oriented as shown in the diagram. The block ml moves on a surface with coefficient of kinetic friction μ1, and the coefficient of static friction between the two blocks is μ2. What is the minimum force F which must be applied to ml such that m2 remains stationary relative to m1? = M2 F FF m2 ml M² (Block 21 Fir² = m₂a Fix/ma 9= F= 12+ mgm, ma Block 1: F-F-F=m, a F-m,gu₁₂ = F12/m2 12 m.g Fimm, gu,arrow_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