Part A nsider the heaviest box of 150 lb that you can push at stant speed across a level floor, where the coefficient of etic friction is 0.45, and estimate the maximum horizontal ce that you can apply to the box. A box sits on a ramp that is lined at an angle of 60.0° above the horizontal. The efficient of kinetic friction between the box and the ramp is If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.) Express your answer in pounds. 5. m = lb Submit Request Answer

Part A nsider the heaviest box of 150 lb that you can push at stant speed across a level floor, where the coefficient of etic friction is 0.45, and estimate the maximum horizontal ce that you can apply to the box. A box sits on a ramp that is lined at an angle of 60.0° above the horizontal. The efficient of kinetic friction between the box and the ramp is If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.) Express your answer in pounds. 5. m = lb Submit Request Answer

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Description: Having trouble with this question. Part AConsider the heaviest box of 150 lb that you can push atconstant speed across a level floor, where the coefficient ofkinetic friction is 0.45, and estimate the maximum horizontalforce that you can apply to the

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

Problem 111AP: A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0with the horizontal. A...

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Check Your Understanding A block of mass 1.0 kg rests on a horizontal surface. The frictional coefficients for the block and surface are s=0.50and k=0.40 . (a) What is the minimum horizontal force required to move the block? (b) What is the block’s acceleration when this force is applied?
Calculate the maximum acceleration of a car that is heading up a 4.00slope (one that makes an angle of 4.00with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On wet concrete. (b) On wet concrete. (C) On ice, assuming that s=0.100 , the same as for shoes on ice.
A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80of that for the static case. (a) What is the coefficient of static friction? (b) What is the maximum force that can be applied upward along the plane on the rope and not move the block? (c) With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? (d) If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? (e) Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?
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Suppose you have a 120-kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood surfaces. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will its acceleration then be? The coefficient of sliding friction is known to be 0.300 for this situation.
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A skydiver will reach a terminal velocity when the air drag equals his or her weight. For a skydiver with a large body, turbulence is a factor at high speeds. The drag force then is approximately proportional to the square of the velocity. Taking the drag force to be FD=12Av2 , and setting this equal to the skydiver's weight, find the terminal speed for a person falling "spread eagle.
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