9th Class Physics · Punjab Board · Chapter 5

Chapter 5: Work, Energy and PowerNumericals Solved Step by Step

All 11 numericals of 9th Class Physics Chapter 5, Work, Energy and Power, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.

Start Full chapter guide MCQs Short Questions Long Questions

Chapter 5· Work, Energy and Power

Numericals Solved Step by Step

Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.

Numerical 5.1

A force of 20 N acting at an angle of 60° to the horizontal is used to pull a box through a distance of 3 m across a floor. How much work is done?

Given Data

F = 20 N
θ = 60°
S = 3 m

To Find

Work done W

Formula

W = FS cos θ

Calculation

W = FS cos θ = 20 × 3 × cos 60°.
= 20 × 3 × 0.5 = 30 J.

Result

W = 30 J

Numerical 5.2

A body moves a distance of 5 m in a straight line under the action of a force of 8 N. If the work done is 20 J, find the angle which the force makes with the direction of motion.

Given Data

S = 5 m
F = 8 N
W = 20 J

To Find

Angle θ

Formula

W = FS cos θ, so cos θ = W ÷ (FS)

Calculation

cos θ = W ÷ (FS) = 20 ÷ (8 × 5) = 20 ÷ 40 = 0.5.
θ = cos⁻¹(0.5) = 60°.

Result

θ = 60°

Numerical 5.3

An engine raises 100 kg of water through a height of 80 m in 25 s. What is the power of the engine? (g = 10 m s⁻²)

Given Data

m = 100 kg
h = 80 m
t = 25 s
g = 10 m s⁻²

To Find

Power P

Formulas

W = mgh

P = W ÷ t

Calculation

Work done = mgh = 100 × 10 × 80 = 80 000 J.
P = W ÷ t = 80 000 ÷ 25 = 3200 W.

Result

P = 3200 W (3.2 kW)

Numerical 5.4

A body of mass 20 kg is at rest. A 40 N force acts on it for 5 s. What is the kinetic energy of the body at the end of this time?

Given Data

m = 20 kg
vᵢ = 0
F = 40 N
t = 5 s

To Find

Kinetic energy Ek

Formulas

a = F ÷ m

vf = vᵢ + at

Ek = ½mv²

Calculation

a = F ÷ m = 40 ÷ 20 = 2 m s⁻².
vf = 0 + 2 × 5 = 10 m s⁻¹.
Ek = ½ × 20 × 10² = 1000 J.

Result

Ek = 1000 J

Numerical 5.5

A girl is swinging on a swing. At the lowest point she is 1.2 m above the ground, and at the highest point she is 2.0 m above the ground. What is her maximum velocity, and where? (g = 10 m s⁻²)

She rises 0.8 m between the lowest and highest points of the swing.

Given Data

lowest point h₁ = 1.2 m
highest point h₂ = 2.0 m
drop h = 2.0 − 1.2 = 0.8 m
g = 10 m s⁻²

To Find

Maximum velocity v, and where it occurs

Formulas

loss of Ep = gain of Ek: mgh = ½mv²

v = √(2gh)

Calculation

At the highest point she is momentarily at rest; going down 0.8 m, her potential energy changes into kinetic energy.
mgh = ½mv², so v² = 2gh = 2 × 10 × 0.8 = 16.
v = √16 = 4 m s⁻¹.

Result

v = 4 m s⁻¹, at the lowest point of the swing

Note: The mass cancels, so it is not needed.

Numerical 5.6

A person pushes a lawn mower with a force of 50 N making an angle of 45° with the horizontal. If the mower is moved through a distance of 20 m, how much work is done?

Given Data

F = 50 N
θ = 45°
S = 20 m

To Find

Work done W

Formula

W = FS cos θ

Calculation

W = 50 × 20 × cos 45° = 1000 × 0.707.
W = 707 J.

Result

W ≈ 707 J

Numerical 5.7

Calculate the work done in (i) pushing a 5 kg box up a frictionless inclined plane 10 m long that makes an angle of 30° with the horizontal (ii) lifting the box vertically from the ground to the top of the inclined plane. (g = 10 m s⁻²)

The slope is 10 m long at 30°, so its top is 10 sin 30° = 5 m high.

Given Data

m = 5 kg
length of slope S = 10 m
θ = 30°
g = 10 m s⁻²
no friction

To Find

(i) Work done up the slope
(ii) Work done lifting straight up

Formulas

force along the slope = mg sin θ

W = F × S

h = S sin θ

W = mgh

Calculation

(i) Force needed along the slope = mg sin 30° = 5 × 10 × 0.5 = 25 N.
Work = 25 × 10 = 250 J.
(ii) Height of the top: h = 10 × sin 30° = 5 m.
Work = mgh = 5 × 10 × 5 = 250 J.

Result

(i) 250 J (ii) 250 J

Note: The work is the same either way: a frictionless slope needs a smaller force, but over a longer distance.

Numerical 5.8

A box of mass 10 kg is pushed up a ramp 15 m long with a force of 80 N. If the box rises through a height of 5 m, what is the efficiency of the system? (g = 10 m s⁻²)

An 80 N push along a 15 m ramp raises the box 5 m.

Given Data

m = 10 kg
S = 15 m
F = 80 N
h = 5 m
g = 10 m s⁻²

To Find

Efficiency

Formulas

input = F × S

useful output = mgh

efficiency = output ÷ input × 100%

Calculation

Input energy = F × S = 80 × 15 = 1200 J.
Useful output = mgh = 10 × 10 × 5 = 500 J.
Efficiency = 500 ÷ 1200 × 100 = 41.7%.

Result

Efficiency ≈ 41.7%

Note: The other 700 J is used against friction on the ramp.

Numerical 5.9

A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power.

Given Data

F = 600 N
S = 5 m
t = 15 s

To Find

Power P

Formulas

W = F × S

P = W ÷ t

Calculation

W = 600 × 5 = 3000 J.
P = 3000 ÷ 15 = 200 W.

Result

P = 200 W

Numerical 5.10

A 40 kg boy runs up stairs 10 m high in 8 s. What power does he develop? (g = 10 m s⁻²)

Given Data

m = 40 kg
h = 10 m
t = 8 s
g = 10 m s⁻²

To Find

Power P

Formula

P = mgh ÷ t

Calculation

Work = mgh = 40 × 10 × 10 = 4000 J.
P = 4000 ÷ 8 = 500 W.

Result

P = 500 W

Numerical 5.11

A force F acts through a distance L on a body. The force is then increased to 2F, which acts through a further distance 2L. Sketch a force-displacement graph and calculate the total work done.

The area under the graph is the work: FL for the first part, 2F × 2L for the second.