9th Class Physics Unit No. 6 Work and Energy (Numerical Problems with Solution)
UNIT No. 6
Work and Energy
(Numerical Problems with Solutions)
Distance
covered by the cart = S = 35 m
Force on the
cart = F = 300 N
Work done by
the man = W = ?
Formula
W=FS
Solution
W=FS
Putting values in this equation
W= (300 N)(35 m)
F= 10500 J Ans
6.2 A block weighing 20 N is lifted 6 m vertically upward. Calculate the potential energy stored in it.
Weight of the
block = w = mg = 20 N
Block is lifted
to a height = h = 6 m
potential
energy stored in the block = P.E = ?
Formula
P.E = mgh
Solution
P.E = mgh
Putting values in this equation
P.E = (20 N)(6 m)
P.E = 120 J Ans
6.3 A car weighing 12 kN has speed of 20 ms-1 . Find its kinetic energy.
Weight of the
car = w = 12 k N = 12x103
N = 12000 N
Speed of car =
v = 20 ms-1
Kinetic energy
of the car = K.E = ?
Formula
\[K.E = \frac{1}{2}mv^{2}\]
w = mg
Solution
First, we will find the mass of the car
w = mg
\[m = \frac{w}{g}\]
Putting values in this equation
\[m = \frac{12000}{10}\]
m = 1200 kg
Now
\[K.E = \frac{1}{2}mv^{2}\]
Putting values in this equation
\[K.E = \frac{1}{2}(1200)(20)^{2}\]
K.E = 240000 J
K.E = 240 kJ Ans
6.4 A 500 g stone is thrown up with a velocity of 15ms-1
. Find its
(i) P.E. at its maximum height
(ii) K.E. when it hits the ground
mass of the
stone = m = 500 g = 0.5 kg
velocity of
stone = v = 15 ms-1
gravitational
constant = g = 10 ms-1
potential
energy of stone at its maximum height = P.E = ?
Kinetic energy
of the stone when in hits the ground = K.E = ?
Formula
P.E = mgh\[K.E = \frac{1}{2}mv^{2}\]
\[K.E = \frac{1}{2}mv^{2}\]
Solution
Potential
Energy of stone at its maximum height (P.E)
P.E = mgh
---------------- (1)
1st
we will find the maximum height “h” that stone will reach, using 3rd
equation of motion
\[2aS = v_{f}^{2} - v_{i}^{2}\]
Here
Initial velocity = vi =
15 ms-1
Initial velocity = vf =
0
acceleration = a = acceleration due to gravity = g = 10
(As the stone is
moving upward therefore value of g will be negative i.e., a = - g = - 10 ms-1
)
distance = S = maximum height reached by the stone = h = ?
Putting values in 3rd equation of motion
\[2aS = (0))^{2}- (15)^{2}\]
2(-10)h = 0 – 225
-20h = -225
h = 11.25 m
putting values in equation (1)
P.E =
(0.5)(10)(11.25)
P.E = 56.25 J Ans
Kinetic
Energy of stone when it hits the ground (K.E)
\[K.E = \frac{1}{2}mv^{2}\] --------------- (2)
1st
we will find the velocity with which the stone will hit the ground using 3rd
equation of motion
\[2aS = v_{f}^{2}- v_{i}^{2}\]
Here
Initial velocity = vi =
0
Initial velocity = vf = ?
acceleration = a = acceleration due to gravity = g = 10
distance = S = maximum height reached by the stone = h = 11.25 m
Putting values in 3rd equation of motion
\[2gS = v_{f}^{2}- (0))^{2}\]
\[2(10)(11.25) = v_{f}^{2}- (0))^{2}\]
\[2(10)(11.25) = v_{f}^{2}\]
\[225 = v_{f}^{2}\]
\[v_{f}^{2}= \sqrt{225}\]
vf = 15 ms-1
Putting values in this equation
\[K.E = \frac{1}{2}mv^{2}\]
\[K.E = \frac{1}{2}(0.5)(15)^{2}\]
K.E = 56.25 J Ans
Potential Energy
of stone at its maximum height = P.E = 56.25 J
Kinetic Energy of the stone when in hits the ground = 56.25 J
6.5 On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 ms-1. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg.
Height of the
slope = h = 6 m
Speed of
cyclist = v = 1.5 ms-1
mass of the
cyclist and his bicycle = m = 40 kg
gravitational
constant = g = 10 ms-1
Kinetic energy
of the cyclist = K.E = ?
potential
energy of cyclist = P.E = ?
Formula
\[K.E = \frac{1}{2}mv^{2}\]
P.E = mgh
Solution
Kinetic
Energy of stone when it hits the ground (K.E)
\[K.E = \frac{1}{2}mv^{2}\]
Putting values in this equation
\[K.E = \frac{1}{2}(40)(1.5)^{2}\]
\[K.E = \frac{1}{2}(40)(2.25)\]
K.E = 45 J Ans
Potential
Energy of stone at its maximum height (P.E)
P.E = mgh
Putting values
in this equation
P.E = (40)(10)(6)
P.E = mgh
P.E = 2400 J Ans
6.6 A motorboat moves at a steady speed of 4 ms-1
. Water resistance acting on it is 4000 N. Calculate the power of its engine.
Speed of
motorboat = v = 4 ms-1
Water
resistance acting on the boat = F = 4000 N
Power
of the engine of boat = P = ?
Formula
\[P = \frac{W}{t}\]
W = FS
Solution
\[P = \frac{W}{t}\]
As
W =FS
Putting this value in above equation
\[P = \frac{FS}{t}\]
\[P = F(\frac{S}{t})\
P = Fv
Putting values
in this equation
P = (4000)(4)
P = 16000 W
P = 16x103
W
P = 16 kW Ans
6.7 A man pulls a block with a force of 300 N through 50
m in 60 s. Find the power used by him to pull the block.
Force applied
by the man = F = 300 N
Distance
covered by the block = 50 m
Time
of application of force = t = 60 s
Power
applied by the man = P = ?
Formula
\[P = \frac{W}{t}\]
W = FS
Solution
\[P = \frac{W}{t}\]
As
W =FS
Putting this value in above equation
\[P = \frac{FS}{t}\]
Putting values in this equation
\[P = \frac{FS}{t}\]
\[P = \frac{(300)(50)}{60}\]
P = 250 W Ans
6.8 A 50 kg man moved 25 steps up in 20 seconds. Find his
power if each step is 16 cm
high.
Mass of the man
= m = 50 kg
Number of steps
= N = 25
Height of each
step = 𝛥h = 16 cm =
0.16 m
height=h=(Number
of steps)(height of each step)
height=h=(N)( 𝛥h)
height=h=(25)( 0.16)
height=h= 4 m
Time
= t = 20 s
Power
of the man = P = ?
Formula
\[P = \frac{W}{t}\]
W = FS
Solution
\[P = \frac{W}{t}\]
As
W =FS = wh = mgh
Putting this value in above equation
\[P = \frac{mgh}{t}\]
Putting values in this equation
\[P = \frac{(50)(10)(4)}{20}\]
P = 100 W Ans
6.9 Calculate the power of a pump which can lift 200 kg of
water through a height of
6 m in 10 seconds.
Mass of water =
m = 200 kg
height=h= 6 m
Time
= t = 10 s
Power
of the man = P = ?
Formula
\[P = \frac{W}{t}\]
W = FS
Solution
\[P = \frac{W}{t}\]
As
W =FS = wh = mgh
Putting this value in above equation
\[P = \frac{mgh}{t}\]
Putting values in this equation
\[P = \frac{(200)(10)(6)}{10}\]
P = 1200 W Ans
6.10
An electric motor of 1hp is used to run water pump.
The water pump takes 10 minutes to fill an overhead tank. The tank has a
capacity of 800 litres and height of 15 m. Find the actual work done by the
electric motor to fill the tank. Also find the efficiency of the system.
(Density of water = 1000 kgm-3)
(Mass of 1 litre of water = 1 kg)
Data
Power of electric motor = 1hp = 746 W
Time taken by the pump to fill the tank = t = 10
minutes = (10)(60s) = 600 s
Capacity of tank to store water in kilograms = m = 800 kg
Height of the tank = 15 m
Actual work done by the motor to fill the tank = W = ?
Efficiency of the system = ?
Formula
\[P = \frac{W}{t}\]
\[%efficiency = \frac{required form of output}{total input energy}X100\]
Solution
Actual work done by the motor to fill the tank
\[P = \frac{W}{t}\]
W = P x t
Putting values in
this equation
W = 746 x 600
W = 447600 J Ans
Efficiency of the system
\[%efficiency = \frac{required form of output}{total input energy}X100\]
Here
Input energy =
actual work done by the motor = W = 447600 J
Output work = mgh =
(800)(10)(15) = 120000 J
Putting values in the above equation
\[%efficiency = \frac{447600}{120000}X100\]
% efficiency = 26.8 % Ans
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