FSC 1st Year Physics Chapter 4 Exercise & Numericals Notes

In this post, I am sharing FSC 1st Year Physics Chapter 4 Solved Exercise & Numericals Notes PDF for the students of Intermediate Part 1. This chapter’s name is work and energy and this is the 4th chapter in the 1st year physics textbook. Students can download the 11th class chapter 4 Work and Energy Notes PDF on their laptop or mobile. They can also read these PDF notes online. These 11th Physics Notes are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. I have already posted the complete Class 11 Physics Notes PDF chapter-wise.

FSC Part 1 Physics Chapter 4 Work and Energy Solved Exercise Numericals Notes PDF

A man pushes a lawn mower with a 40N force directed at angle of 20⁰ downward from the horizontal. Find the work done by the man as he cuts a strip of grass 20m long.

Solution:
Data;

                        Force exerted = F = 40N
                        length of strip of grass = 20m
                        Angle = q = 20⁰
to find:
                        work = W =?
Calculation:
                        we know that
                        W =F.d
                        = Fd cosq
                        = 40(20) cos20⁰
                        W = 800 x 0.93
                        W = 7.51 x 10²J

A car of mass 800kg travelling at 54kmh-1 was brought to rest in 60 meters. Find the average retarding force on the car. What has happened to the original kinetic energy?

Solution:
Data;
                        mass of the car = m = 800 kg
                        initial velocity = v_i = 54kmh^-1
                        = v_i = 54 x 1000/ 60 x60
                        initial velocity = v_i = 15ms^-1
                        final velocity = v_f = 0
                        distance covered by car = d = 60
To find:
            average retarding force = F = ?
            What happened to the original kinetic energy =?
Calculation;
according to the work energy principle
                        Fd = 1/2mv_f² – 1/2mv_i²
                        Fx60 = 1/2x800x(0)²– ½(800)(15)
                        F = 0- 400 x 225/60
                        F = -1500N

A 1000kg automobile at the top of an incline 10 meter high and 100m long is released and rolls down the hill. What is its speed at the bottom of the incline of the average retarding force due to friction is 480N?

Solution:
Data;
                        mass of automobile = m = 1000kg
                                                            f = 480N
                        length = l = 100m
to find :
                        final velocity = v = ?
calculation;
            we know that
                        K.E =1/2 mv²
                        P.E = mgh
according to the law of conversion of energy
loss of P.E = work done against friction = gain in K.E
so we write
            mgh = f x l = 1/2mv²
            substituting the value we have
  1000 x 9.8 x 10 – 480 x 100 = 1/2 x 1000 x v²
98000 – 48000 = 500v²
100 =v²
v = 10 ms^-1

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Related Links:

1. 1st Year Physics Chapter 1 Solved Numericals Notes PDF
2. FSC 1st Year Physics Chapter 2 Solved Numericals Notes PDF
3. FSC 1st Year Physics Chapter 3 Numericals Notes