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NCERT Solution Chapter 1 Some Basic Concepts of Chemistry NCERT Solution Chapter 2 Structure of Atom NCERT Solution Chapter 3 Classification of Elements and periodicity in properties NCERT Solution Chapter 4 Chemical Bonding and Molecular Structure NCERT Solution Chapter 5 States of Matter NCERT Solution Chapter 6 Thermodynamics NCERT Solution Chapter 7 Equilibrium NCERT Solution Chapter 8 Redox Reaction NCERT Solution Chapter 9 Hydrogen NCERT Solution Chapter 10 The S-Block Element NCERT Solution Chapter 11 The P-Block Element NCERT Solution Chapter 12 Organic Chemistry- Some Basic Principle and Techniques NCERT Solution Chapter 13 Hydrocarbons NCERT Solution Chapter 14 Environmental Chemistry

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NCERT Solution  Chapter 1 Sets NCERT Solution  Chapter 2 Relations & Functions   NCERT Solution Chapter 3 Trigonometric Functions NCERT Solutions Chapter 4 Principle of Mathematical Induction NCERT Solution Chapter 5 Complex Numbers and Quadratic Equations NCERT Solution Chapter 6 Linear Inequalities NCERT Solution Chapter 7 Permutations and Combinations NCERT Solution Chapter 8 Binomial Theorem NCERT Solution Chapter 9 Sequence and Series NCERT Solution Chapter 10 Straight Lines NCERT Solution Chapter 11 Conic Sections NCERT Solution Chapter 12 Introduction to Three dimensional Geometry NCERT Solution Chapter 13 Limits and Derivatives NCERT Solution Chapter 14 Mathematical Reasoning NCERT Solution Chapter 15 Statistics NCERT Solution Chapter 16 Probability Free Download NCERT Solutions Class 11 Physics PDF

Hello! Here is another problem from the Giancoli book. Chapter 6, # 25.

A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

Alright, here are the givens:

m = 285-kg

d = 22.0 m

a = 0.160g

initial velocity = 0

Here I have drawn a little diagram:

PART A - Finding the tension (FT) in the cable.

We have values for mass and acceleration. We can use the net force and what we know about the force of gravity on the load to find the tension in the cable.

F(net) = ma = FT - Fg

Here, we are subtracting the force of gravity because it is in the downward direction and in this case downward is negative.

We know that Fg = mg therefore,

ma = FT - mg

FT = ma + mg

FT = (285kg)(0.160g) + (285kg)g

FT = (285kg)(0.160)(9.8 m/s2) + (285kg)(9.8 m/s2)

FT = 3.24 x 103 N

Alright! We are done with Part A.

PART B - Finding the Net Work on the Load

We know the net force. We can use that and the distance that the load has traveled because of that force to find the net work. Here, the net force is parallel to the displacement, and we can use it to find the net work.

F(net)= ma = 446.88 N

W(net) = F(net)d

W(net) = (446.88 N)(22.0 m) = 9.83 x 103 J

Yay! We have found the answer to part B!

PART C - Finding the Work Done by the Cable on the Load

To find the work done by the cable only, we need to use the force that the cable applied on the load. This is the force of tension on the cable. We know that the force of tension on the cable is 3.24 x 103 N. We are going to use the value that wasn't rounded. If you did the calculation, you should have a number 3239.88 N.

WT = FTd = (3239.88 N)(22.0m) = 7.13 x 104 J

That was simple! Now, onto part D.

PART D - Finding the Work Done by Gravity on the Load

The work done by gravity on the load is simply the force of gravity on the load multiplied by the displacement. In this case, the work done is negative because the force is working against the movement of the load.

Wg = -Fgd = - mgd = - (285kg)(9.8m/s2)(22.0 m) = -6.14 x 104 J

PART E - Finding the Final Speed of the Load

We know that the initial velocity is zero.

We use the Work-Energy Principle

Wnet = KEf - KEi

Wnet = 0.5mvf2 - 0.5mvi2

Wnet = 0.5mvf2

We re-write the equation for v.

v = √(Wnet / 0.5 m) = √(9831.36 J / 142.5 kg) = 8.31 m/s