udent/courseuserassignments/view?courseuserassignment-34634576&mode-; This print
ID: 3280546 • Letter: U
Question
udent/courseuserassignments/view?courseuserassignment-34634576&mode-; This print-out should have 6 questions. what is the new acceleration of the box? Multiple-choice questions may continue on Answer in units of m/s the next oolumn or page find all choices before answering. questions may continue on Answer in units of m/s 004 (part 1 of 2) 10.0 points A dock worker loading crates on a ship finds that a 15 kg crate, initially at rest on a hori- 001 10.0 points A box of books weighing 337 N is shoved zontal surface, requires a 78 N horizontal force across the floor of an apartment by a force ofto set it in motion. However, after the crate 582 N exerted downward at an angle of 25.3 is in motion, a horizontal force of 51 N is below the horizontal The acceleration of gravity is 9.8 N required to keep it moving with a constant speed If the coefficient of kinetic friction between The acceleration of gravity is 9.8 m/s? box and floor is 0.583, how long does it take Find the coefficient of static friction be to move the box 7.18 m, starting from rest?tween crate and floor. Answer in units of s. 005 (part 2 of 2) 10.0 points Find the coefficient of kinetic friction. 002 (part 1 of 2) 10.0 points A student decides to move a box of books into her dormitory room by pulling on a rope at- tached to the box. She pulls with a force of 006 10.0 points The coeflicient of static friction between the 160 N at an angle of 29 above the horizon 4.26 kg crate and the 28.3 incline is 0.339. tal. The box has a mass of 26.2 kg, and theThe acceleration of gravity is 9.8 m/s coefficient of friction between box and floor is 0.297 The acceleration of gravity is 9.8 m/s 29 28.3 26.2 kg = 0.297 Find the acceleration of the box. What minimum force F must be applied to the erate perpendicular to the incline to pre- vent the crate from sliding down the incline? Answer in units of m/s Answer in units of N 003 (part 2 of 2) 10.0 points The student now starts moving the box up a 14.1 incline, keeping her 160 N force directed 29 If the coeficient of friction is unchanged,Explanation / Answer
001]
Net Force on the book =
Fcos25.3 - ukR = Fnet
=> 582cos25.3 - 0.583[mg + Fsin25.3] = ma
=> 582cos25.3 - 0.583[337 + 582sin25.3] = (337/9.8)a
=> a = 5.371 m/s2
S = 7.18 m
u = 0 m/s
So, S = ut + (1/2)at2
7.18 = 0 + (1/2)(5.371)t2
=> t = 1.6351 s.
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