Class 9 Physics Chapter 5 numerical problems are according to the new syllabus of the Punjab Board. Chapter 5 is related to Work, Energy & Power, in which you will study the force, motion, and field effects of gravitation.
Class 9 Physics Chapter 5 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
Work Done = F.S
W = F × S × cos(θ)
Power = Work / Time
F = ma
vf = vi + at
K.E. = ½mv²
P.E. = mgh
Efficiency = (Work output/Work input) × 100
Numerical Problems Chapter 5 – Gravitation
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?
Solution:
Force (F) = 20 N
Angle (θ) = 60° to horizontal
Distance (S) = 3 m
Work Done (W) = ?
Using Formula,
Work Done = F.S
W = F × S × cos(θ)
W = 20 × 3 × cos(60°)
W = 20 × 3 × 0.5
W = 30 Joules
5.2 A body moves a distance of 5 metres in a straight line under the action of a force of 8 newtons. If the work done is 20 Joules, find the angle which the force makes with the direction of motion of the body.
Solution:
Distance (S) = 5 m
Force (F) = 8 N
Work done (W) = 20 J
Angle (θ) = ?
Using Formula,
Work Done = F.S
W = F × S × cos(θ)
20 = 8 × 5 × cos(θ)
cos(θ) = 20/(8 × 5)
cos(θ) = 0.5
θ = cos-1(0.5)
θ = 60°
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?
Solution:
Mass (m) = 100 kg
Height (h) = 80 m
Time (t) = 25 s
g = 9.8 m/s²
Work done (W) = ?
Power (P) = ?
Using Formula,
Work done = mgh
W = 100 × 9.8 × 80
W = 78,400 J
Using Formula,
Power = Work / Time
P = W/T
P = 78,400/25
P = 3,200 Watts
5.4 A body of mass 20 kg is at rest. A 40 N force acts on it for 5 seconds. What is the kinetic energy of the body at the end of this time?
Solution:
Mass (m) = 20 kg
Force (F) = 40 N
Time (t) = 5 s
K.E. = ?
Step 1: Find acceleration
F = ma
a = F/m
a = 40/20
a = 2 m/s²
Step 2: Find final velocity
vf = vi + at
vf = 0 + 2 × 5
vf = 10 m/s
Step 3: Calculate kinetic energy
K.E. = ½mv²
K.E. = ½ × 20 × 10²
K.E. = 1000 Joules
5.5 A ball of mass 160 g is thrown vertically upward. The ball reaches a height of 20 m. Find the potential energy gained by the ball at this height.
Solution:
Mass (m) = 160 g = 0.16 kg
Height (h) = 20 m
g = 9.8 m/s²
P.E. = ?
Using formula,
Potential Energy = mgh
P.E. = 0.16 × 9.8 × 20
P.E. = 31.36
P.E. ≈ 32 J
5.6 A 0.14 kg ball is thrown vertically upward with an initial velocity of 35 m s¹. Find the maximum height reached by the ball.
Solution:
Mass (m) = 0.14 kg
Initial velocity (u) = 35 m/s
g = 9.8 m/s²
Height (h) = ?
Using conservation of energy:
Initial KE = Final PE
½mv² = mgh
½v² = gh
½ × (35)² = g × h
h = (35)²/(2 × 9.8)
h = 1,225/19.6
h = 61.25 m
5.7 A girl is swinging on a swing. At the lowest point of her swing, she is 1.2 m from the ground, and at the highest point she is 2.0 m from the ground. What is her maximum velocity and where?
Solution:
Lowest point height = 1.2 m
Highest point height = 2.0 m
Height difference = 0.8 m
g = 9.8 m/s²
v = ?
Using conservation of energy:
PE at highest = KE at lowest
mgh = ½mv²
gh = ½v²
2gh = v²
Taking square root
v = √(2gh)
v = √(2 × 9.8 × 0.8)
v = √15.68
v = 4 m/s
5.8 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?
Solution:
Force (F) = 50 N
Angle (θ) = 45°
Distance (S) = 20 m
Work done (W) = ?
Using formula,
Work done = F × S × cos(θ)
W = 50 × 20 × cos(45°)
W = 50 × 20 × 0.707
W = 707 J
5.9 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 up from the ground to the top of the inclined plane.
Solution:
Mass (m) = 5 kg
Height (h) = 10 m
Angle (θ) = 30°
g = 9.8 m/s²
Work done (W) = ?
Using formula,
Work done = w×h
W = mgh sin(θ)
W = 5 × 9.8 × 10 × sin(30°)
W = 5 × 9.8 × 10 × 0.5
W = 245 J
W ≈ 250 J
(ii) Lifting box vertically
Height = 10 × sin(30°) = 5 m
Work done = mgh
W = 5 × 9.8 × 5
W = 245 J
W ≈ 250 J
5.10 A box of mass 10 kg is pushed up along a ramp 15 m long with a force of 80 N. If the box rises up a height of 5 m, what is the efficiency of the system?
Solution:
Mass (m) = 10 kg
Ramp length (S) = 15 m
Height gained (h) = 5 m
Applied force (F) = 80 N
Efficiency = ?
Work output = mgh
Wo = 10 × 9.8 × 5
Wo = 490 J
Work input = F × S
Wi = 80 × 15
Wi = 1200 J
Efficiency = (Work output/Work input) × 100
E = (490/1200) × 100
E = 41.7%
5.11 A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power.
Solution:
Force (F) = 600 N
Distance (S) = 5 m
Time (t) = 15 s
Power (P) = ?
Step 1: Calculate work done
Work = F × S
W = 600 × 5
W = 3000 J
Step 2: Calculate power
Power = Work/Time
P = 3000/15
P = 200 Watts
5.12 A 40 kg boy runs up-stair 10 m high in 8 s. What power he developed.
Solution:
Mass (m) = 40 kg
Height (h) = 10 m
Time (t) = 8 s
g = 9.8 m/s²
Power (P) = ?
Step 1: Calculate work done against gravity
Work = mgh
W = 40 × 9.8 × 10
W = 3920 J
Step 2: Calculate power
Power = Work/Time
P = 3920/8
P = 490 W
P ≈ 500 W
5.13 A force F acts through a distance L on a body. The force is then increased to 2F that further acts through 2L. Sketch a force displacement graph and calculate the total work done.
Solution:
Initial force F through distance L
Then force 2F through distance 2L
Step 1: Calculate work in first part
W₁ = F × L
Step 2: Calculate work in second part
W₂ = 2F × 2L
W₂ = 4FL
Step 3: Calculate total work
Total Work = W₁ + W₂
W = FL + 4FL
W = 5FL or 5 units
Class 9 Physics Full Book Numerical Problems with Solution
