StudentAidUnit 5: Work, Energy and Power — Numericals
9th Class Physics · Unit 5: Work, Energy and Power
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
Force
F = 20 N
Angle
θ = 60^circ
Distance
S = 3 m
Formula
Work formula
W = FScosθ
Working
W = (20)(3)cos 60^circ
Evaluate cosine
W = (60)(0.5)
Result
Work done
W = 30 J
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.
Given
Distance
S = 5 m
Force
F = 8 N
Work
W = 20 J
Formula
Work formula rearranged
W = FScosθ therefore cosθ = WFS
Working
θ = cos-1left(WFSright)
θ = cos-1left(208 × 5right)
θ = cos-1left(2040right)
θ = cos-1(0.5)
Result
Angle
θ = 60^circ
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?
Given
Mass
m = 100 kg
Height
h = 80 m
Time
t = 25 s
Formula
Power formula
P = mght where g = 10 m s-2
Working
P = (100)(10)(80)25 = 8000025
Result
Power
P = 3200 W
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?
Given
Mass
m = 20 kg
Initial velocity
vi = 0 m s-1
Time
t = 5 s
Force
F = 40 N
Formula
Acceleration
a = Fm
Working
a = 4020 = 2 m s-2
Formula
Final velocity
vf = vi + at
vf = 0 + (2)(5) = 10 m s-1
Kinetic energy
K.E = 12mvf2
Working
K.E = 12(20)(10)2
K.E = (10)(100)
Result
Kinetic energy
K.E = 1000 J
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.
Given
Mass
m = 160 g = 1601000 kg = 0.16 kg
Height
h = 20 m
Formula
Potential energy
P.E = mgh where g = 10 m s-2
Working
P.E = (0.16)(10)(20)
Result
Potential energy
P.E = 32 J
5.6.A 0.14 kg ball is thrown vertically upward with an initial velocity of 35 m s^{-1}. Find the maximum height reached by the ball.
Given
Mass
m = 0.14 kg
Initial velocity
Vi = 35 m s-1
Final velocity
Vf = 0 m s-1
Formula
Kinematic equation
2gh = Vf2 - Vi2 where g = -10 m s-2
Working
2(-10)h = (0)2 - (35)2
(-20)h = -1225
h = -1225-20
Result
Maximum height
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?
Given
Highest point
h1 = 2.0 m
Lowest point
h2 = 1.2 m
Change in height
Δh = h1 - h2 = 2.0 - 1.2 = 0.8 m
Formula
Energy at highest point
At highest: P.E = mgh, K.E = 0
Energy at lowest point
At lowest: P.E = 0, K.E = 12mv2
Conservation of energy
mgh + 0 = 0 + 12mv2
Simplify
gh = 12v2
2gh = v2
V = sqrt{2gh} = sqrt{2(10)(0.8)} = sqrt{16}
Result
Maximum velocity
V = 4 m s-1
Location of maximum velocity
At the lowest position
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?
Given
Force
F = 50 N
Angle
θ = 45^circ
Distance
S = 20 m
Formula
Work formula
W = FScosθ
Working
W = (50)(20)cos(45^circ)
W = (1000)(0.7071)
Result
Work done
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.
Given
Mass
m = 5 kg
Weight
W = mg = (5)(10) = 50 N, g = 10 m s-2
Distance along plane
S = 10 m
Angle
θ = 30^circ
Formula
(i) Work on inclined plane
W = FSsinθ
Working
(i)
W = (5)(10)(10)sin(30^circ)
W = 500(0.5)
Result
(i) Work done on incline
W = 250 J
Formula
(ii) Work lifting vertically
W = mgh
(ii) Using h = S\sin\theta
h = 10sin(30^circ) = 10(0.5) = 5 m
Working
W = (5)(10)(5)
Result
(ii) Work done lifting vertically
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?
Given
Mass
m = 10 kg
Distance
S = 15 m
Force
F = 80 N
Height
h = 5 m
Formula
Work input
W = FS
Working
W = (80)(15) = 1200 J
Formula
Useful work (potential energy)
P.E = mgh
Working
P.E = (10)(10)(5) = 500 J
Formula
Efficiency
Efficiency = P.EW × 100
Working
%Efficiency = 5001200 × 100
Result
Efficiency
= 41.7%
5.11.A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power.
Given
Force
F = 600 N
Distance
S = 5 m
Time
t = 15 s
Formula
Work done
W = FS
Power
P = Wt = FSt
Working
P = (600)(5)15 = 300015
Result
Power
P = 200 W
5.12.A 40 kg boy runs up-stairs 10 m high in 8 s. What power he developed.
Given
Mass
m = 40 kg
Height
h = 10 m
Time
t = 8 s
Formula
Potential energy gained
W = P.E = mgh
Power
P = mght where g = 10 m s-2
Working
P = (40)(10)(10)8 = 40008
Result
Power developed
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.
Given
First segment force
F1 = F
First segment distance
s1 = L
Second segment force
F2 = 2F
Second segment distance
s2 = 2L
Formula
Work in first segment
W1 = F × L
Work in second segment
W2 = 2F × 2L = 4FL
Total work
W = W1 + W2
Working
W = (F × L) + (4F × L) = FL + 4FL
Result
Total work done
W = 5FL