Solve it. At a construction site, a 60.0 kg bucket of concretehangs from a light (but strong) cable that passes over alight friction-free pulley and is connected to an 80.0 kgbox on a horizontal roof (see the figure (Figure 1)). Thecable pulls horizontally on the box, and a 50.0 kg bagof gravel rests on top of the box. The coefficients offriction between the box and roof are shown. Thesystem is not moving.FigureGravelBoxMs = 0.700Mk0.400=1 of 1ConcretePart CSuddenly a worker picks up the bag of gravel. Use energy conservation to findthe speed of the bucket after it has descended 2.40 m from rest. (You can checkyour answer by solving this problem using Newton's laws.)Express your answer in meters per second.VE ΑΣΦv=6.86Submit=Previous Answers Request AnswerReview | ConstantsX Incorrect; Try Again; 3 attempts remainingm/s

Given data: M=60 kgm1=80 kgm2=50 kg Need to determine the speed speed of the bucket after descended…

At a construction site, a 60.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 80.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 50.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving. Figure Gravel Box Mg = 0.700 Mk= 0.400 1 of 1 Concrete Part C Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.40 m from rest. (You can check your answer by solving this problem using Newton's laws.) Express your answer in meters per second. ΨΕ ΑΣΦ v= 6.86 Submit Previous Answers Request Answer ? Review | Constants X Incorrect; Try Again; 3 attempts remaining m/s

At a construction site, a 60.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 80.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 50.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving. Figure Gravel Box Mg = 0.700 Mk= 0.400 1 of 1 Concrete Part C Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.40 m from rest. (You can check your answer by solving this problem using Newton's laws.) Express your answer in meters per second. ΨΕ ΑΣΦ v= 6.86 Submit Previous Answers Request Answer ? Review | Constants X Incorrect; Try Again; 3 attempts remaining m/s

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

Solve it.

At a construction site, a 60.0 kg bucket of concrete
hangs from a light (but strong) cable that passes over a
light friction-free pulley and is connected to an 80.0 kg
box on a horizontal roof (see the figure (Figure 1)). The
cable pulls horizontally on the box, and a 50.0 kg bag
of gravel rests on top of the box. The coefficients of
friction between the box and roof are shown. The
system is not moving.
Figure
Gravel
Box
Ms = 0.700
Mk
0.400
=
1 of 1
Concrete
Part C
Suddenly a worker picks up the bag of gravel. Use energy conservation to find
the speed of the bucket after it has descended 2.40 m from rest. (You can check
your answer by solving this problem using Newton's laws.)
Express your answer in meters per second.
VE ΑΣΦ
v=6.86
Submit
=
Previous Answers Request Answer
Review | Constants
X Incorrect; Try Again; 3 attempts remaining
m/s

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